Electrostatics Content 1. Introduction of Electrostatics 2. Properties of charges 3. Coulomb’s Law 4. Electric Field 5. Electric Filed lines and its properties 6. Electric potential energy Equipotential Curve 7. Gauss’s Law and its application 8. Properties of conductor 9. Electric Dipole 10. Capacitor 11. Van de Graaff Generator
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Electrostatics
1. Introduction of Electrostatics Electrostatics is the branch of Physics, which deals with static electric charges or charges at rest or slow-moving. Electrostatics: Properties of Stationary Charge Magnetism: Properties of Moving charge
2. Properties of charge “Object is charged to indicate that it has a charge imbalance.” Electric Charge is an intrinsic property of matter (like mass) which causes it to experience a force when near other electrically charged matter. There are two types of charge, positive and negative. Electric Charge SI Symbol Q or q Si units Coulomb Other units e Derivation from other quantities Q It Electron
e
Negative
1.6 10 19 C
Proton Neutron
p n
Positive Neutral
1.6 10 19 C Zero
When a glass rod is rubbed with silk, the transfer of electrons takes place. Glass rod loses some electrons and hence it becomes positively charged and silk gains those electrons and it becomes negatively charged. Electron Transfer
+
e _
Thus, there are two types of charges namely positive charge and negative charge. Note: Charge produces Electric field (Stationary Charge) and Magnetic field (Moving Charge), and radiates energy. “Charge can be transferred from one part of the system to another system, but net charge will have a constant value” or “charge can never be created nor be destroyed” Properties of electric charge: 1. Like charges repel each other whereas unlike charges attract each other. 2. Law of conservation of charge: The algebraic sum of total charges on a system is always constant. 3. It is electron which is responsible for charging of a body and not the proton. 4. The charge on a body cannot be a fraction of electron charge. (Charge is quantized)
q ne
5. Charge is always associated with mass. In Charging, the mass of a body changes. If electrons are removed from the body, the mass of the body will decreases and the body will becomes positively charged. On the other hand, if electrons are added to a body, the mass of the body will increase and the body will acquire a net negative. 6. Accelerated charge radiates energy.
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Electrostatics Problem 1: Which one charge is not possible on a body? A. 8 10 20 C Solution: (A). n
B. 2.4 10 19 C
C. 16 10 19 C
D. 3.2 10 19 C
q 0.8 10 19 1 ; which must be an integer value, Charge is always transferring in the e 1.6 10 19 2
integral multiple of the charge of electron. (B). n
q 2.4 10 19 3 e 1.6 10 19 2
q 16 10 19 10 e 1.6 10 19 q 3.2 10 19 (D). n 2 e 1.6 10 19 (C). n
Hence, A, B are not possible and C, D is possible. For the point of view of electrostatic, there are two type of materials; Conductors and Insulators In conductor electric charges free to move, the outer electrons of each atom or molecule are only weakly bound to it. These electrons are almost free to move throughout the body of the material and called free electrons (conduction electrons). When such a materials is placed in an electric field, the free electrons move in a direction opposite to the field. Such materials are called conductors. (Example: Metal, human body and Earth) Insulator, in which all the electrons are tightly, bound their respective atoms or molecule. There are no free electrons. When such material is placed in an electric field, the electrons may slightly shift opposite to the field but they cannot leave their parent atoms and hence can’t move through long distance. Such materials are called dielectrics. (Example: Glass, rubber and plastic) Methods of charging conductors i. By rubbing ii. By conduction (By contact) iii. By Induction (From a distance) 1. By rubbing (Frictional electricity): When a glass rod is rubbed with a silk cloth the glass rod acquires some positive charge and the silk cloth acquire negative charge by the same amount. Comb is passed through dry hair. Cloud also becomes charged by friction. 2. By conduction (Charging by contact): After conduction charge becomes equally Distributed between A & B, if A and B are exactly same in shape, size material and finishing, Otherwise unequally.
Uncharged
B
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Gets Charged ++ + + q + + ++ + + + + A +
++++ +
+
q/2 A
+ + + + +
Charged
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q/2
A
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Electrostatics 3. By Induction (From a distance): It is also possible to charge a conductor in a way that does not involve contact. A positively charged rod brought close to metal sphere. In the sphere free electrons close to the rod move close to this side.
+ + + + Insulated Rod
+ -
Grounding wire + + - + - + + - + + + - + + - + + + + + + Earth-
- -
-
-
-
-
-
-
-
Problem 2: Three small identical balls have charges 3 10 12 C , 8 10 12 C and 4 10 12 C respectively. They are brought in contact and then separated. Calculate (i) charge on each ball (ii) number of electrons in excess or deficit on each ball after contact. Solution: (i) The charge on each ball
q
q1 q 2 q3 3 8 4 4 4 10 3 10 C 3 3
(ii) Since the charge is positive, there is a shortage of electrons on each ball.
q 3 10 4 1.875 10 7 19 e 1.6 10 ∴ Number of electrons =1.875 107 . n
Problem 3: Two spherical conductors B and C having equal radii and carrying equal charges on them repel each other with a force F when kept apart at some distance. A third spherical conductor having same radii as that of B but uncharged is brought in contact with B, then brought in contact with C and finally removed away from both. Find the new force of repulsion between B and C. Solution: Initially, Force between B and C is F
KQ 2 R2
B
Q
C
Q
Q=0
A
R A third uncharged sphere ‘A’, is brought in contact with B, (Charge transfer will take place by conduction)
B
A
C
Q
R
Q/2 Q/2
Amount of transferred Charge is depend upon the shape of conductor, material.
Then third charged sphere with charge Q/2, is brought in contact with C, (Again charge transfer take place till equal charge)
Q
B
C
A
R Q/2
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Q/2
3Q/4 C
B R Q/2
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A
3Q/4
Electrostatics Here, Total Charge is conserved. Total Charge on C and A is Q Q / 2 3Q / 2 , which equally distributed on the both sphere, Charge on each sphere is 3Q / 4 . Finally;
B
C R
3Q/4
Q/2 Finally Force between B and C
F'
K Q / 23Q / 4 3 KQ 2 3 F R2 8 R2 8
3. Coulomb’s Law: r
q1
q2
“The force between two point charges is directly proportional to product of their magnitude q1 & q2 , and inversely proportional to the square of the distance ‘r’ between them. The direction of forces is along the line joining the two point charges.”
F q1q2
F
…………………….
1 r2
(i)
……………………
(ii)
Combined two equations
F k
q1q 2 r2
Where k is a constant k 9 10 9 Nm 2 / C 2
F k
Vector Form:
1 4 0
and 0 = permittivity of the free space.
q1 q 2 ^ (r ) r2
F k
q1 q 2 (r ) r3
Important Points Regarding Coulomb’s law 1. Charges are assumed to be rest. 2. Charges are assumed to be point particle. 3. Magnitude of the force
F k
q1q 2 r2
4. Direction of the force between two charges: Force between two positive charges
F Q
F
Q
Repulsive Force
Force between two opposite charges Q
F
F
Q
F
Force between two negative charges Q
F Q
Repulsive Force
Attractive Force
(i). “Same charges push each other in the opposite direction (Force is repulsive) and opposite charges pull each other in the direction of towards each other (Force is attractive) and this force always along the straight line joined both charges.”
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Electrostatics (ii).
If force is negative force is attractive If force is positive force is repulsive
(iii). Force on the both charge is equal in magnitude (iv). There is no any force exerted on any point charge by self. 5. Effect of the medium on the force between two charges: When a dielectric medium is completely filled in between charges rearrangement of the charges inside the dielectric medium takes place and the force between the same tow charge decreases by a factor of K known as dielectric constant or specific inductive capacity (SIC) of the medium, K is also called relative permittivity of the medium (relative means with respect to free space).
F
Hence in the presence of the medium:
q1q 2 4 0 r r 2 1
r K = Dielectric constant/ relative permittivity of the medium. F
q1
F/K
q2
q1
In Vacuum
In Medium
q2
Relative Permittivity of the medium: Dielectric constant is the ratio of the force of attraction or repulsion between the two similar point charges in the air to the ratio of the force of attraction or repulsion between point charges separated by same distance in the medium.
K
Fair FMedium
6. Principle of Superposition of electric force: Total force on a given charges is the vector sum of all the individual force exerted on it by all other charges, each individual force being calculated by Coulomb’s law.
F1 F12 F13 F14 F15 ...... q3 q1
q
q5 q2
q4
7. Force on a test charge due to continuous distribution of charges. qo r
dq q
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F
q0 4 0
dq ^ r 2 .r
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Electrostatics 8. Charge densities: Linear Charge Density: Charge per unit length,
Q C l m
Surface Charge density: Charge per unit area
Q C A m2
Volume Charge Density: Charge per unit volume
Q C V m3
Problem 4: Two point charges A and B separated by a distance R attract each other with a force of 12 10 3 N . Find the force between A and B when the charges on them are doubled and distance is halved. Solution: Given that F
KQ 2 12 103 N 2 R Force between two opposite charges Q
F
F
Q
Attractive Force
Then, the charges on them are doubled and distance is halved, Force between two opposite charges 2Q
New Force is F '
F
’
F
’
2Q
K (2Q)(2Q) KQ 2 16 ( R / 2) 2 R2
Magnitude of new force is16 F 16 12 103 0.192 N . Problem 5: If a charge q is placed at the centre of the line joining two equal charges Q such that the system is in equilibrium then find the value of q. Solution: System is equilibrium means that net force on each particle is zero. q
Q
Q
Force on middle charge is always zero, Force on charge located at left hand side or right hand side,
Fnet Fq FQ
KQq KQ 2 0 ( r ) 2 ( 2r ) 2
By solving this q
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Q 4
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Electrostatics Problem 6: Two small identical balls P and Q, each of mass
3 gm , carry identical charges and are suspended by 10
threads of equal lengths. At equilibrium, they position themselves as shown in figure. What charge on each ball?
600
600
Solution: Free body diagram Tsin600
T
0
60
600
Fe
Tcos600 mg
T sin 600 mg
kq2 T cos 60 2 l 0
With the help of these two equations
tan 600
mgl 2 mgl 2 q 10 7 C kq2 k tan 600
For tension in the string
kq2 T (mg ) 2 l 2
Problem 7: Point charges having values 0.1C , 0.2C , 0.3C and 0.2C are placed at the corners A, B, C and D respectively of square of side one meter. Calculate the magnitude of the force on a charge of 1C placed at centre of the square. Solution: 0.2C B
0.1C A AC 2 AB 2 BC 2 1 1 2 AC 2 1 1 AO 2m m CO BO DO 2 2
1C O
(0.1106 )(1106 )
FA 9 109
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1 2
FA
FB
0.2C D
Where
1m
2
FD
F
FC
C 0.3C
0.0018 N (along OC)
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Electrostatics FC 9 109
6
(0.3 10 )(110 ) 1 2
FB 9 109
Where
FD 9 109
6
0.0054 N
2
(0.2 106 )(1106 ) 1 2
2
(0.2 106 )(1106 ) 1 2
2
(Along OC)
0.0036 N (Along OD)
0.0036 N
(Along OD)
Force on charge at O due to A (Repulsive) and due to C is (attractive) along the AC
F1 FA FC 0.0018 0.0054 0.0072 N Similarly force on charge due to B and D is along OD
F1 FB FD 0.0036 0.0036 0.0072 N There are two forces equal in magnitude and acting perpendicular Net force F
F12 F22 2F1F2 cos 900 2 F 1.414 0.0072 0.01018N
Alternate Method: Use vector form. Problem 8: Three charges q1 ,q2 and q3 are placed as shown in the figure. The x-component of the force on
q1 is proportional to ‒q3
y
a
A.
q 2 q3 sin b2 a2
B.
C.
q 2 q3 sin b2 a2
D.
Ѳ b X
‒q1
+q2
q 2 q3 cos b2 a2
q 2 q3 cos b2 a2
Solution: Answer is C ‒q3
X component of the force
kq q F1 12 2 b
a
Fx F1 F2 sin
‒q1
b
+q2
F2 sin F2 cos
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kq1 q 2 kq1 q3 2 sin b2 a q q Fx 22 32 sin b a
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F2
Fx
kq1 q 3 a2
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Electrostatics Problem Set 1 1. What is the Coulomb’s law of electrostatics? Explain the dependency of the medium. Write law in the vector form. 2. Define the dielectric constant of the medium? 3. Two point charges of 2C and 6C repel each other with a force of 12 N. If each is given an additional charge of 4C , what will be the new force? 4. Explain: Two identical metallic spheres of exactly equal masses are taken. One is given a positive charge q coulombs and the other an equal negative charge. Are their masses after charging equal? 5. Two point charges each of q C separated by 1m distance experience a force of F. How much force is experienced by them if they are immersed in water, keeping distance of separation between them same. (Dielectric constant for water K = 80). 6. Force of attraction between two point charges placed at a distance d is F. What distance apart should they be kept in the same medium so that force between them is F/3? 7. Does Coulomb’s law of electric force obey the Newton’s law? 8. How does the force between two point charges change if the dielectric constant of the medium in which they are kept increases? 9. What is quantization of electric charge? 10. Two charged spherical conductors, each of radius R, are distant d ( d 2R ). They carry charges + q and – q. Will the force of attraction between them be exactly
q2 4 0 d 2
?
11. Two identical metallic spheres, having unequal opposite charges are placed at a distance 0.9 m apart in air. After bringing them in contact with each other they are again placed at the same distance apart. Now the force of repulsion between them is 0.025 N. Calculate the finale charge on each of them. 12. In what ways does a charge differs from mass? 13. Two similar and equal charged identical metal spheres A and B repel each other with a force of 2 105 N . A third identical uncharged sphere C is touched with A and then placed at the midpoint between A and B. Calculate the net electric force on C. 14. Three point charges of 2C,3C and 3C are kept at the vertices A, B, C respectively of an equilateral triangle of side 20 cm. What should be the sign and the magnitude of the charge to be placed at the midpoint (M) of side BC so that the charge at A remains in equilibrium?
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Electrostatics 15. Two point charges of values q and 2q are kept at a distance d apart from each other in air. A third charge Q is to be kept along the same line in such a way that net force acting on q and 2q is zero. Calculate the position of charge Q in terms of q and d. 4. Electric Field A field of force surrounding a charged particle. Electric field strength: - The strength of electric field at a point is defined as the force experienced by one coulomb of positive charge placed at that point. It is a vector quantity and its direction is the direction in which the force acts on positive charge placed at that point.
F E Limit q0 0 q 0 Where, F is the force on charge q 0 due to electric field due to charge Q. Hence, electric field due to a point charge Q is given by
E
1 Q 4 0 r 2
Electric filed due to system of n charges (use vector addition)
E1 E12 E13 E14 E15 ...... Electric field due to continuous distribution of charge (use integration)
E dE P r
dq
E
q
1
dq ^ .r 4 0 r 2
Problem 9: An infinite number of charges, each q coulomb, are placed along x – axis at x 1m,3m,9m,...... Calculate the electric field at the point x 0 due to these charges, if (i). All the charges are same sign (ii) consecutive charges have opposite signs. Solution: (i). E E1 E3 E9 ..........
x=0
E
q
q
x=1
x=3
q
∞
x=9
kq kq kq ......... 12 32 92
1 1 1 1 E kq 2 2 2 ....... kq 1 1 1 3 9 9 a (Sum of infinite G.P.; S , Where a is first term and r is common ratio) 1 r q 9 E NC 1 4 0 8
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Electrostatics (ii). E E1 E3 E9 .......... q x=0
x=3
x=1
E
q
-q
∞
x=9
kq kq kq ......... 12 32 92
1 1 1 1 E kq 2 2 2 ....... kq 1 1 1 3 9 9 q 9 E NC 1 4 0 10
Problem 10: Point charges having values 1C,5C and 2C are placed at the corners A, B and C respectively of an equilateral triangle of side 2 m in free space. Determine the magnitude of intensity at the point D midway between A and C. B ‒5 µC
EB A 1 µC
Solution:
EC
D
EA
C 2 µC
106 The intensity E A at D due to charge at A is given by E A 9 10 2 1 9
1 1 AD 2 AC 2 2m 1m EA 9 103 N / C ; Along DC
The intensity EC at D due to charge at C is given by
2 106 N /C EC 18 103 N / C 2 1 The magnitude of the resultant of EC and ED is given by
EC 9 109
(Long DA)
EC EA (18 103 9 103 ) N / C 9 103 N / C The intensity EB at D due to charge at B is given by
5 106 N / C 15 103 N / C Along DB 3 BD BD 3m In right angled triangle cos 300 2 EB 9 109
If E is the magnitude of the intensity, then E (9 103 ) 2 (15 103 ) 2 1.749 104 N / C
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Electrostatics Problem 11: Find the electric field at a perpendicular distance r from the midpoint of the straight wire of length l and has charge Q . Solution: Charge density is
Q l
Let’s take a small line element length of dx and has charge dq at a distance x from the centre of the straight wire. Electric field at the point due to small charge element R is given by
E
kdq (RP ) 2
(Along the RP); where r OP
And RP OR2 OP2 x 2 r 2
E
kdq x r2 2
Electric field at point P due to the similar point charge at S
E
kdq x r2 2
Vertical component due to both point charge has cancelled and resultant electric field due to the point charges is given by dE 2E cos E sin dx
R
E
x l
Ө x Ө
O
E
dx
S
2E cos
P
E sin
So, net electric field due to straight wire l/2
E2 0
Where dq
Q dx and cos l l/2
E2 0
kdq cos x r2 2
x x2 r 2 l/2
kdq kQr 2kQr cos 2 dx 2 2 2 2 3/ 2 x r l(x r ) l 0
l/2 2kQr 1 x 2kQ 1 E l r 2 x 2 r 2 0 l r
l/2
(x 0
2
dx r 2 )3 / 2
l/2
2kQ 1 l l r l 2 4r 2 l / 22 r 2 0 l/2
2kQ 1 2 r l 4r 2 dx I 2 2 3/ 2 (x r ) E
Let x r tan
dx r sec2
r sec2 d cos d sin 1 x I 3 3 r sec r2 r2 r2 x2 r2
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Electrostatics Important Point: Electric field due to infinite length wire
2kQ 1 2kQ E Limit Limit l 0 l 0 r l 2 4r 2 rl 2k E r
1 2k Limit 2 l 0 r r 2 4 l
1 2 r 2 4 l
(Alternate method: use Gauss’s Law) 5. ELECTRIC FIELD LINES/ ELECTRIC FORCE LINES: Graphical Representation Electric filed in a region can be graphical represented by drawing certain curves known as lines of electric force or electric field lines. Electric field provides a means for visualizing the direction and magnitude of electric field. Electric field lines
+Q
The following points should be remembered: 1. Lines of force diverge out from a positive charge and converge at a negative charge or Electric field lines extend away from positive charge (where they originate) and towards negative charge (where they terminate) 2. The tangent to a line of force at any point gives the direction of electric field at that point. 3. There is no physical existence of electric field lines. 4. Lines must lie along the radii. 5. Lines of force never intersect. 6. In a uniform field, the filed lines are straight parallel and uniformly spaced. 7. Electric field line always flow from higher potential to lower potential. 8. Filed lines never exist inside a conductor and start or end normally from the surface of a conductor. 9. Electric field lines also give the indication of equipotential surfaces. 10. The number of lines of force originating from or ending on a charge is proportional to the magnitude of the charge. Important figure:
─
+
─
+
Positive Charge
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Negative Charge
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Electrostatics
+ ─
+
Positive Charge
Positive Charge
-Q Electric field is zero everywhere inside B. a metal (conductor). Field lines do not enter a metal plus these are perpendicular to metal surface.
B
A.
E A EB FORCE ON A CHARGE q IN ELECTRIC FIELD
F qE
F qE
Force on positive charge always in the direction of electric field, and force on negative charge opposite to electric field direction. Problem 12: A uniform electric field E is created between two parallel, charged plates as shown in figure (29-W9). An electron enters the field symmetrically between the plates with of speed u 0.The length of each plate is l . Find the angle of deviation of the path of the electron as it comes out of the field.
l +
+
+
+ +
+
+
+
+
E
Ө
v0 ‒
‒
‒
‒
‒
‒
Solution: The acceleration of the electron is a
‒
‒
‒
eE in the upward direction. The horizontal velocity remain u0 as m
there is no acceleration in this direction. Thus, the time taken in crossing the field is
t
l u0
… (1)
The upward component of the velocity of the electron as it emerges from the field region is
u y at
eEl mu0
The horizontal component of the velocity remains
ux u0 .
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Electrostatics The angle ɵ made by the resultant velocity with the original direction is given by
tan
uy ux
eEl 2 mu0
Thus, the electron deviates by an angle
tan 1
eEl 2 mu0
6. Electric potential energy: The electrostatic potential energy of a system of point charges is the work required to assemble this system of charges by bringing them in form an infinite distance. Electrostatic Potential energy of the system of two particles: r2
Q O
P
dr R
r
r1
S
Assumptions: First, we assume that the test charge q is so small that it does not disturb the original configuration, namely the charge Q at the origin (or else, we keep Q fixed at the origin by some unspecified force). Second, in bringing the charge q from R to P, we apply an external force Fext just enough to counter the repulsive electric force FE (i.e,
Fext FE ). This means there is no net force on or acceleration of the charge q when it is brought from R to P, i.e., it is brought with infinitesimally slow constant speed. Work done by the external force is the negative of the work done by the electric force, and gets fully stored in the form of potential energy of the charge q. Thus, work done by external forces in moving a charge q from R to P is P
P
WRP Fext dr FE dr R
R
This work done is against electrostatic repulsive force and gets stored as potential energy. 1 1 1 1 kQq 1 1 dr kQq dr kQq kQq 2 2 r r r r r1 r2 R r2 2
r
r
P
WRP
If point R at the infinite then r2
U
kQq r1
Electrostatic Potential: Electric potential is a property of an electric field, regardless of whether a charged object has been placed in that field; it is measured in joules per coulomb, or volts. The Electric potential at a point in an electric field is the energy required (Work to be done by an external force) per unit charge to bring a small positive charge from infinity to that point without any acceleration to it.” Electric potential due to a point charge r Q
P
VP
W kQ q r
Note: i. Most important is potential difference.
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Electrostatics ii. The potential at a point due to a positive source charge is positive and that due to negative source charge is negative. Relationship between E and V: f
1. If electric field is given
P
V E . dl
V E . dl
i
Where
E Exi E y j Ez k
dl dx i dy j dz k dV
dV
dV
2. If potential is given then E i j k dy dz dx 3. Potential due to a group of charges
V V1 V2 V3 ............. 4. Potential due to continuous charge distribution
V dV Electric potential energy: - “Potential Energy of a system of point charges is defined as the amount of work done to assemble this system by bringing them in from an infinite distance.” Potential Energy of two point charges
U
1 q1 q 2 4 0 r
Change in potential energy of a point charge in moving it from one point A to another B in fixed uniform field:
U q(VB VA ) W Equipotential Surface:
Equipotential Surface - -----------
Equipotential Surface
Electric filed lines
Properties of Equipotential Surface: i. There is no electric field in any direction lying on or along the equipotential surface. (Because there is no potential difference) ii. Electric field and hence lines of force are always perpendicular to the Equipotential surface. iii. No work is required to be done in moving a charge from one point to another on an equipotential surface.
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Electrostatics Problem 13: Calculate (i) the potential at a point due a charge of 4 107 C located at 0.09m away (ii) work done in bringing a charge of 2 109 C from infinity to the point. Solution: (i) The potential due to the charge q1 at a point is V
1
q1 4 0 r
4 107 V 9 10 4 104 Volt 0.09
9
(ii) Work done in bringing a charge q2 from infinity to the point is W = q2 V = 2 × 10 −9 × 4 × 104 W = 8 × 10 −5 J Problem 14: Calculate the electric potential at a point P, located at the centre of the square of point charges shown in the figure.
q2 24nC
q1 12nC
d =1.3m P q4 17nC
q3 31nC Solution: Potential at a point P is V The distance
r
Total charge
1 q1 q2 q3 q4 4 0 r r r r
d 1.3 0.919m 2 2 = q1 q2 q3 q4
(12 24 31 17) 109 C 9 = 36 10 C 36 109 C 9 ∴ V 9 10 = 352.6 V 0.919 =
Problem 15: Three charges – 2 × 10−9 C, +3 × 10 −9C, – 4 × 10 −9C are placed at the vertices of an equilateral triangle ABC of side 20 cm. Calculate the work done in shifting the charges A, B and C to P, Q and R respectively which are the mid points of the sides of the triangle.
A
q1 2nC
P
q2 3nC
B
R
Q
C
q3 4nC
Solution: The potential energy of the system of charges, U
1 q1q2 q2q3 q3q1 4 0 r r r
Work done in displacing the charges from A, B and C to P, Q and R respectively
W U f Ui
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Electrostatics U i and U f are the initial and final potential energy of the system.
2 109 3 109 3 109 (4 109 ) 4 109 (2 109 ) U i 9 109 4.5 10 7 J 0.02 0.02 0.02 2 109 3 109 3 109 (4 109 ) 4 109 (2 109 ) U i 9 109 9 10 7 J 0.01 0.02 0.01 ∴ Work done = 4.5 10 7 J Problem 16: An alpha particle with kinetic energy 10 Me V is heading towards a stationary tin nucleus of atomic 50. Calculate the distance of closest approach. Solution: In approaching the nucleus, kinetic energy of alpha particle is converted into electrical potential energy, so if r is the distance of closest approach:
1 2 1 q1q2 1 q1q2 mv r 2 4 0 r 4 0 ( K .E.) 1 (2e)(50e) 4 0 ( KM )
r
r (9 109 )
r 14.4 1015 m 14.4 fm
(2 1.6 1019 )(1.6 1019 50) (10 106 1.6 109 )
Passage for Q 17 to 19: Two fixed charges – 2Q and Q are located at the points with coordinates ( - 3a, 0) and ( + 3a, 0) respectively in the x-y plane. Given that all points in the x- y plane where the electric potential due to the two charges is zero, lie in the circle. 17. Radius of the given circle A. 3a
B. 4a
C. 5a
D. 6a
18. The expression of potential V (x) at a general point on the x axis is
1
2
1
2
A. B. 3a x 3a x 3a x 3a x
1
3
C. 2a x 2a x
1
3
D. 2a x 2a x
19. If a particle of charge +q starts form rest at the centre of the circle, by a short quantitative argument that the particle eventually crosses the circle, its speed when it does so 1/ 2
1/ 2
1/ 2
1 Qq 1 Qq 1 2Qq A. B. C. D. 4 0 2ma 4 0 ma 4 0 ma Solution: Let P( x, y) be general point on x – y plane. Electric potential at point P would be
V VQ V2Q
V
1 Q 4 0 ( x 3a) 2 y 2
1 4 0
2Q ( x 3a) 2 y 2
……………………………….(1)
Given that potential is zero
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1 4 0
Qq 4ma
1/ 2
Electrostatics
4 (3a x) y (3a x) y ( x 5a) y (4a) 2
2
2
2
2
2
2
P(x, y)
-2Q
Q
(3a,0)
(3a,0)
This equation of a circle of radius 4a and centre at (5a,0)
P(x, y) V=0 0 (a,0)
(5a,0)
(9a,0)
(b). On x –axis potential will be undefined (or say )at x 3a and x 3a , because charge Q and 2Q are placed at these two points. So, between 3a x 3a we can find potential by putting y 0 in equation (1). Therefore,
Q 1 2 for 3a x 3a 4 0 3a x 3a x V 0 at x a V at x 3a And V at x 3a For x 3a , there is again a point where potential will become zero so for x 3a , we can write: Q 1 2 V for x 3a 4 0 x 3a 3a x For x 3a , we can write Q 1 2 V for x 3a 4 0 x 3a x 3a Potential at centre at x 5a will be, Q 1 2 Q V = positive 4 0 2a 8a 4 0 a V
Potential on the circle is zero. Since, potential > potential on the circumference on it, the particle will cross the circle because positive charge moves from higher potential to lower potential. Speed of the particle, while crossing the circle would be,
v
2qV qQ m 8 0 ma
Here, V is potential difference between the centre and circumference of the circle.
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Electrostatics Problem Set 2 1. In an electric field an electron is kept freely. If the electron is replaced by a proton, what will be the relationship between the force experienced by them? 2. Two point charges of 3C each are 100 cm apart. At what point on the line joining the charges will the electric intensity be zero? 3. What is the signs of q1 and q2 ?
q2
q1 4. Draw lines of force to represent a uniform electric field. 5. Sketch the line of force due to two equal positive point charges placed near each other? 6. Consider three charged rods A, B and C. A and B repel each other while A and C attract each other. What will be the nature of force between B and C? 7. Two point electric charges of unknown magnitude and sign are placed a distance‘d’ apart. The electric field intensity is zero at a point, not between the charges but on the line joining them. Write two essential conditions for this to happen. 8. Two point charges of 5 1019 C and 20 1019 C are separated by a distance 2 m. Find the point on the line joining them at which electric field intensity is zero. 9. What is an equipotential surface? Write down any tow properties of the equipotential surface. 10. What is the work done in moving a change of 10 nC between two points on an equipotential surface? 11. Draw an equipotential surface in a uniform in a uniform electric filed. 12. Draw an equipotential surface for a point change Q 0. 13. A positive change is moved in an electrostatic filed from a point at high potential to a point at low potential. How does its kinetic energy and potential energy change? 14. Three concentric metal spheres A, B and C have radii R1 , R 2 and R 3 respectively and have change Q1 , Q 2 and Q 3 . What is the potential and intensity of electric field at a point P between the spheres
C and B at a distance r from the centre O of the spheres? 15. The work done in moving a charge of 3 C between two points is 6 J . what is the potential difference between the points? 16. Explain what is meant by an electric line of force? Give its 4 important properties? Explain with sketches.
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Electrostatics
7. Gauss’ law Electric Flux of an electric field through a surface: The electric flux through a Gaussian surface is proportional to the net number of electric field lines passing through that surface. Or Electric Flux: - The electric flux linked with the surface are the product of the surface area and the component of electric intensity taken perpendicular to the surface of the field.
E . dS Direction of surface is perpendicular to the surface pointing outwards. The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by 0 .
qin
E . dS
0
^
^
Problem 20: The electric field in a region is given by E p i q j , where p & q are constants. Find the net flux passing through a circle area of radius of r parallel to (i). x – y plane
^
^
(ii). y – z plane (iii). z – x plane. ^
Solution: (i). E p i q j , area vector parallel to x – y plane is S r 2 k
^
^
^
Flux E . dS ( p i q j ) r 2 k 0
^
^
^
(ii). E p i q j , area vector parallel to z – y plane is S r 2 i
^
^
^
Flux E . dS ( p i q j ) r 2 i pr 2
^
^
^
(iii). E p i q j , area vector parallel to z – x plane is S r 2 j
^
^
^
Flux E . dS ( p i q j ) r 2 j qr 2 Applications of Gauss’s law: 1. Charged Conductor: An electric conductor has a large number of free electrons and when placed in an electric field, these electrons redistribute themselves to make the field zero at all the points inside the conductor. If a charge is injected anywhere in the conductor, it must come over to the surface of the conductor so that the interior is always charge free. Also, if the conductor has a cavity, the charge must come over to the surface. Earthing a conductor All conductor which are not given any external charge, are also very nearly at the same potential. The potential of the earth is often taken to be zero. If a conductor is connected to the earth, the potential of the conductor becomes equal to that of the earth, i.e. zero. If the conductor was at some other potential, charge will flow from it to earth or from earth to it to br ing its potential to zero.
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Electrostatics + ¯ -q¯
+ +
¯
Conductor
+ ¯
¯ +
¯ -q¯
+ .+q
¯ ¯ ¯
.+qR
+ ¯
Conductor
+ ¯
+
¯
The potential due to the charge at the centre V
¯ ¯
.+qR
¯
¯C 2+
¯
¯
¯
kq r
kq r kq The potential due to the charge at the outer surface V r kq Net potential = V r The potential due to the charge at the inner surface V
After earthing, the charge at outer surface flows to earth and potential of the sphere becomes zero.
2. Electric field due to a uniformly charged Sphere: Here is a spherical geometry, where the charges are evenly distributed throughout the volume. If the total charge in the sphere is Q, and the sphere has a radius R, then the volume charge density is
By
symmetry,
the
E
Q 4 R 3 3 field
C/m 3
is
everywhere
radial
from
the
center
of
the
(i). If point is inside the Sphere: Use a spherical Gaussian surface,
which is perpendicular to E everywhere. The area vector S is parallel to
E , and the total area is 4r 2 so when the Gaussian surface radius is r R , then qin E . dS
0
E
r 3 0
E
Q
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4 3 R 3
R S
4 0 E dA 0 E 4r 2 r 3 3
E r
r 1 Qr 3 0 4 0 R 3
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sphere.
Electrostatics (ii). If point is outside the Sphere: Use a spherical Gaussian surface,
which is perpendicular to E everywhere. The area vector S is parallel
to E , and the total area is 4r 2 so when the Gaussian surface radius is r R , then qin E . dS
0
4 3
0 E dA 0 E 4r 2 R 3
E
R 3 0 r 2
E
1
S
E
r R
3
Q 4 0 r 2
(iii). Electric field on the surface: Put r R , in any result
E
1
Q 4 0 R 2
Potential due to Solid sphere Potential at an outside point: V Potential on the surface: V
kq r
kq R
Potential at the inside point: E
kqr R3
Potential: V
kq 3 r 2 R 2 2R 2
3. Field due to an infinite long straight charged wire: Consider a long charge with linear charge density . Calculate the electric field at a point P which is at a distance x from the line charge.
The electric flu x ( ) through curved surface =
l
Eds E (2rl) (∵
00 and
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0
the surface area of the curved part is
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E
Electrostatics
2k r
Note: The line charge is assumed infinitely long because this alone would ensure the direction of electric field radially outward at all points around it. 4. Electric field due to a plane Sheet of Charge: Consider a large plane sheet of charge with surface charge density (Charge per unit area) .
E
2 0
Does not depend upon the distance from the charge sheet. (Class Notes) 5. Electric field due to a thick charged conducting plate:
E
0
Electric field inside the conductor E 0 6. The electric field outside a spherical shell of charge with radius R and total charge q is directly radially and has magnitude Outside point
E
Inside Point
E= 0
kq r2
kq r
rR
Potential: V
r VC >VD
B. VA = VB < VC < VD
C. VA < VB < VC < VD
D. VA = VB = VC = VD
7. The separation between the plates of parallel plate capacitor is d and area of each plate is A, when a slab of material of dielectric constant K and thickness t is introduced between the plates, the capacitance becomes: A.
0 A
B.
1 d t 1 K
0 A 1 d t 1 K
C.
0 A 1 d t 1 K
D.
0 A 1 d t 1 K
8. A charge ‘Q’ is distributed over two concentric hollow spheres of radii ‘r’ and ‘R’ (>r) such that surface densities are equal, then potential at the common centre A.
KQ R
B.
KQ r
C.
KQ( R r ) (R 2 r 2 )
D. zero
9. The figure shows two identical parallel plate capacitors connected to a battery with the switch S closed. The switch is now opened and the free space between the plates of the capacitors is filled with a dielectric of dielectric constant (or relative permittivity) 3. Then, the ratio of the total electrostatic energy stored in both capacitors before and after the introduction of the dielectric
S V
A
C B
A.3/5
B. 4/3
C. 4/5
D. None
C
10. To obtain 3F capacity from three capacitors of 2F each, they will be arranged A. All the three in series B. All the three in parallel C. Two capacitors in series and third in parallel with the combination of first two D. Two capacitors in parallel and the third in series with the combination of first two 11. In bringing an electron towards another electron, electrostatic potential energy of the system: A. decreases B. Increases C. remain unchanged D. become zero
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Electrostatics 12. The equivalent capacitance between points A and B of circuit shown in the figure is: B. 1F
A. 9F C. 4.5F
3F A
D.
6F
B
3F
3F
13. Two conductors of radius R1 and R2 capacitances C1 and C 2 have charges q1 and q 2 respectively when they are joined together by a conducting wire, charge redistributes in these conductors q '1 and q ' 2 respectively, then
q '1 q' 2 R A. 1 R2
R B. 2 R1
C.
R D. 1 R2
R1 R2
2
14. Consider a parallel plate capacitor with plate area A. A charge positive q is given to one plate and negative charge q to other plate. Then force with which both plates attract each other is (A = area of the plate) A.
q2 A 0
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q2 2 A 0
Electrostatics Assignment 4 1. A glass rod rubbed with silk acquires a charge of 8 10 12 C . The number of electrons it has gained or lost A. 5 10 7 (gained) B. 5 10 7 (lost) C. 2 10 8 (lost) D. 8 10 12 (lost) 2. The electrostatic force between two point charges kept at a distance d apart, in a medium εr = 6, is 0.3 N. The force between them at the same separation in vacuum is A. 20 N B. 0.5 N C. 1.8 N D. 2 N 3. Electric field intensity is 400 V/m at a distance of 2 m from a point charge. It will be 100 V/ m at a distance? A. 50 cm B. 4 cm C. 4 m D. 1.5 m 4. Two point charges +4q and +q are placed 30 cm apart. At what point on the line joining them the electric field is zero? A. 15 cm from the charge q B. 7.5 cm from the charge q C. 20 cm from the charge 4q D. 5 cm from the charge q 5 A dipole is placed in a uniform electric field with its axis parallel to the field. It experiences A. only a net force B. only a torque C. both a net force and torque D. Neither a net force nor a torque 6. If a point lies at a distance x from the midpoint of the dipole, the electric potential at this point is proportional to A.
1 x2
B.
1 x3
C.
1 x4
D.
1 x3/ 2
7. Four charges +q, +q, −q and –q respectively are placed at the corners A, B, C and D of a square of side a. The electric potential at the centre O of the square is A.
1
q 4 0 a
B.
1
2q 4 0 a
C.
1
4q 4 0 a
D. zero
8. Electric potential energy (U) of two point charges is A.
q1 q 2 4 0 r 2
B.
q1 q 2 4 0 r
C. pE cos
D. pE sin
9. The work done in moving 500 μC charge between two points on equipotential surface is A. zero B. finite positive C. finite negative D. infinite 10 Which of the following quantities is scalar? A. dipole moment B. electric force
C. electric field
D. electric potential
11 The unit of permittivity is A. C 2 N 1m 2
C. H / m
D. NC 2 m 2
B. Nm 2 C 2
12 The number of electric lines of force originating from a charge of 1 C is A. 1.129 × 1011 B. 1.6 × 10 −19 C. 6.25 × 1018
D. 8.85 × 1012
13 The electric field outside the plates of two oppositely charged plane sheets of charge density σ is A.
2 0
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Electrostatics 14. The capacitance of a parallel plate capacitor increases from 5 μf to 60 μf when a dielectric is filled between the plates. The dielectric constant of the dielectric is A. 65 B. 55 C. 12 D. 10 15. A hollow metal ball carrying an electric charge produces no electric field at points A. outside the sphere B. on its surface C. inside the sphere D. at a distance more than twice 16. State Coulomb’s law in electrostatics and represent it in vector form. 17. What is permittivity and relative permittivity? How are they related? 18. Explain the principle of superposition. 19. Define electric field at a point. Give its unit and obtain an expression for the electric field at a point due to a point charge. 20. Write the properties of lines of forces. 21. What is an electric dipole? Define electric dipole moment? 22. Derive an expression for the torque acting on the electric dipole when placed in a uniform field. 23. What does an electric dipole experience when kept in a uniform electric field and non−uniform electric field? 24. Derive an expression for electric field due to an electric dipole (a) at a point on its axial line (b) at a point along the equatorial line. 25. Define electric potential at a point. Is it a scalar or a vector quantity? Obtain an expression for electric potential due to a point charge. 26. Distinguish between electric potential and potential difference. 27. What is an equipotential surface? 28. What is electrostatic potential energy of a system of two point charges? Deduce an expression for it. 29. Derive an expression for electric potential due to an electric dipole. 30. Define electric flux. Give its unit. 31. State Gauss’s law. Applying this, calculate electric field due to (i) an infinitely long straight charge with uniform charge density (ii) an infinite plane sheet of charge of q. 32. What is a capacitor? Define its capacitance. 33. Explain the principle of capacitor. Deduce an expression for the capacitance of the parallel plate capacitor. 34. What is dielectric? Explain the effect of introducing a dielectric slab between the plates of parallel plate capacitor.
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Electrostatics 35. A parallel plate capacitor is connected to a battery. If the dielectric slab of thickness equal to half the plate separation is inserted between the plates what happens to (i) capacitance of the capacitor (ii) electric field between the plates (iii) potential difference between the plates. 36. Deduce an expression for the equivalent capacitance of capacitors connected in series and parallel.
1 Q2 2 37. Prove that the energy stored in a parallel plate capacitor is CV 2 2C 38. What is meant by dielectric polarisation? 39. State the principle and explain the construction and working of Van de Graaff generator. 41. The sum of two point charges is 6 μ C. They attract each other with a force of 0.9 N, when kept 40 cm apart in vacuum. Calculate the charges. 42. Two small charged spheres repel each other with a force of 2 × 10 −3 N. The charge on one sphere is twice that on the other. When one of the charges is moved 10 cm away from the other, the force is 5 × 10 −4 N. Calculate the charges and the initial distance between them. 43. Four charges +q, +2q, +q and –q are placed at the corners of a square. Calculate the electric field at the intersection of the diagonals of the square of side10 cm if q = 5/3 × 10 −9C. 44. Two charges 10 × 10−9 C and 20 × 10 −9C are placed at a distance of 0.3 m apart. Find the potential and intensity at a point mid−way between them. 45. An electric dipole of charges 2 × 10−10C and –2 × 10 −10 C separated by a distance 5 mm, is placed at an angle of 60o to a uniform field of 10Vm−1. Find the (i) magnitude and direction of the force acting on each charge. (ii) Torque exerted by the field 46. An electric dipole of charges 2 × 10−6 C, −2 × 10−6 C are separated by a distance 1 cm. Calculate the electric field due to dipole at a point on its. (i) axial line 1 m from its centre (ii) equatorial line 1 m from its centre. 47. Two charges +q and –3q are separated by a distance of 1 m. At what point in between the charges on its axis is the potential zero? 48. Three charges +1μC, +3μC and –5μ C are kept at the vertices of an equilateral triangle of sides 60 cm. Find the electrostatic potential energy of the system of charges. 49. Two positive charges of 12 μC and 8 μC respectively are 10 cm apart. Find the work done in bringing them 4 cm closer, so that, they are 6 cm apart. 50. Find the electric flux through each face of a hollow cube of side 10 cm, if a charge of 8.85 μC is placed at the centre. 51. A spherical conductor of radius 0.12 m has a charge of 1.6 × 10−7C distributed uniformly on its surface. What is the electric field (i) inside the sphere (ii) on the sphere (iii) at a point 0.18 m from the centre of the sphere? 52. The area of each plate of a parallel plate capacitor is 4 × 10−2 sq m. If the thickness of the dielectric medium between the plates is 10−3 m and the relative permittivity of the dielectric is 7. Find the capacitance of the capacitor.
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Electrostatics 53. Two capacitors of unknown capacitances are connected in series and parallel. If the net capacitances in the two combinations are 6μF and 25μF respectively, find their capacitances. 54. Two capacitances 0.5 μF and 0.75 μF are connected in parallel and the combination to a 110 V battery. Calculate the charge from the source and charge on each capacitor. 55. Three capacitors are connected in parallel to a 100 V battery as shown in figure. What is the total energy stored in the combination of capacitor?
56. A parallel plate capacitor is maintained at some potential difference. A 3 mm thick slab is introduced between the plates. To maintain the plates at the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab. 57. A dielectric of dielectric constant 3 fills three fourth of the space between the plates of a parallel plate capacitor. What percentage of the energy is stored in the dielectric? 58. Find the charges on the capacitor shown in figure and the potential difference across them.
59. Three capacitors each of capacitance 9 pF are connected in series (i) What is the total capacitance of the combination? (ii) What is the potential difference across each capacitor, if the combination is connected to 120 V supply?
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Electrostatics Answer Sheet Problem Set 1 1. Class Notes
2. Class Notes
3. – 4 N (attractive)
4. No their masses not equal.
9. q ne 15.
10. No
11. 1. 5 µ C
5. F/80
6.
7. Yes 8. Force decreases
3d 5
13. 2 10 N 14. 3.93 3N along AM
12. Class Notes
d 2d from charge q or from charge 2q (1 2 ) 1 2
Problem Set 2 1. Class Notes 2. At midpoint 50 cm 3. q1 is negative and q2 is positive 4. Class Notes 5. Class Notes 6. Attractive 7. Class Notes 8. 2/3 m 9. Class Notes 10. Zero 11. Class Notes 12. Class Notes 13. Potential Energy decreases and kinetic energy increases 14. V p
1 Q1 Q2 Q3 4 0 r r R3
E
1 Q1 Q2 4 0 r r 2
15.2 Volt
16. Class Notes
Problem Set 3 1. v – m or N m2 C -1, Scalar
2. q / 2 0
3. (i). 1/3
Assignment 1 1. B 2. A, B, D
4. C
5. A
6. A, B 7. D
5. C
6. B
6. B
7. D
Assignment 2 1. C 2. A 3. B 12. A, B, C, D Assignment 3 1. A 2. A 3. C
3. B
4. A, D
4. A
5. A
(ii). 1/5 th
4, 5, 6: See Class Notes
7. C
8. B
9. B
10. A, B, C
11. A, C
8. C
9. A
10. C
11. B
13. A
12. A
14. D
Assignment 4 1 (b) 2 (c) 3 (c) 4 (c) 5 (d) 6 (a) 7 (d) 8 (b) 9 (a) 10 (d) 11 (a) 12 (a) 13 (d) 14 (c) 15 (c) 35 (i) increases (ii) remains the same (iii) remains the same 41 q1 = 8 × 10−6C , q2 = –2 × 10−6 C 42 q1 = 33.33 × 10−9C, q2 = 66.66 ×10−9 C, x = 0.1 m 43 0.9 × 104 Vm–1 44 V = 1800 V, E = 4000 −1 Vm 45 2 × 10−9N, along the field, τ = 0.866 × 10−11 Nm 46 360 N/C, 180 N C –1 47 x = 0.25 m from +q 48 –0.255 J 49 5.70 J 50 1.67 × 105 Nm2C−1 –1 −1 51 zero, 105 N C , 4.44 × 104 N C 52 2.478 × 10 −9F 53 C1 = 15 μF, C2 = 10μF 54 q = 137.5 μC, q1 = 55 μC, q2 = 82.5 μC 55 0.3 J 56 εr = 5 57 50% 58 q1 = 144 × 10−6C, q2 = 96 × 10 −6C, q3 = 48 × 10 −6C V1 = 72 V, V2 = 48 V 59 3 pF, each one is 40 V
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Electrostatics Introduction of Advance level
Problem: A thin half ring of radius R 20cm is uniformly charged with total charge Q = 0.70 nC . Find the magnitude of the electric field strength at the curvature centre of this half ring. Reference: IE Irodove Solution: Let P and Q are the two small elements whose azimuthal angle is with respect to C.
dq
C
P
Q
d
E
E cos
Electric Field due to small elements P has two components Vertical and horizontal; Similar Elements Q has two components, Horizontal components E sin will cancel. Hence, only Resultant electric field on the center will due to vertical component E cos .
dE E cos
If [90 0 ,90 0 ]
dE 2E cos
If [0,90 0 ]
And E = is the electric field due to small element Q has charge dq
Kdq R2 Q Where; dq d E
SO total electric field due to this ring
K Q E 2 2 d cos 0 R 90
90
E
2 KQ 2 KQ 2Q cos d 2 (1 0) 2 R 0 R 4 0 R 2
E
Q 2 R 2 0 2
Put the values and get
E 0.10kV / m
. A thin non conducting ring of radius R has linear charge density 0 cos , where 0 is a constant, is the azimuthal angle. Find the magnitude of the electric field strength (a). at the centre of the ring; (b). on the axis of the ring as a function of the distance x form its centre. Investigate the obtained function at
x R
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Electrostatics Advance Assignment 1: All questions have only one answer 1. Two vertical metallic plates carrying equal and opposite charges are parallel to each other. A small spherical metallic ball is suspended by a long insulating thread such that it hangs freely in the centre of the two metallic plates. The ball, which is unchanged, is taken slowly towards the positively charged plate and is made to touch the plate. The ball will A. Stick to the positively charged plate B. Come back to the original position and will remain there C. Oscillation between the two plates touching each plate in turn D. Oscillate between the plates without touching them 2. An electron of mass me initially at rest, moves through a certain distance in a uniform electric field in time t1 . A proton of mass m p , also, initially at rest, takes time t 2 to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio t 2 / t1 is nearly equal to:
A. 1
B. m p / me
1/ 2
C. me / m p
1/ 2
D. 1836
3. A charge + q is fixed at each of the points x x0 , x 3x0 , x 5x0 ,..... on the x –axis and charge – q is fixed at each of the points x 2 x0 , x 4 x0 , x 6 x0 ,..... . Here, x 0 is a positive constant. Take the electric potential at a point due to a charge Q at a distance r from it to be Q / 4 0 r . Then the potential at the origin due to the above system of charge is: A. 0
B.
q
C.
8 0 ln 2
D.
q ln 2 4 0 x0
4. Force between two dipoles varies with distance between them A. Inversely with fourth power of the distance B. Inversely with third power of the distance C. Inversely with second power of the distance D. None
5. Four charges equal to – Q are placed at the four corners of a square and a charge q is at its centre . If the system is in equilibrium the value of q is: A.
Q 1 2 2 4
B.
Q 1 2 2 4
C.
Q 1 2 2 2
D.
Q 1 2 2 2
6. If the electric flux entering and leaving an enclosed surface respectively is 1 and 2 , the electric charge inside the surface will be: A. 1 2 0
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Electrostatics 7. At any point on the right bisector of the line joining two equal and opposite charges A. The electric field is zero B. The electric potential is zero C. The electric potential decreases with increasing distance from the centre D. The electric field is perpendicular to the line joining the charge 8. A sheet of aluminum foil of negligible thickness is introduced between the plates of a capacitor. The capacitance of the capacitor: A. Increases B. Decreases C. Remain uncharged D. Becomes infinite 9. A non-conducting ring of radius 0.5 m carries a total charges of 1.11 10 10 C distributed non-uniformly on its l 0
circumference producing an electric field E every where in space. The value of the integral
Edl (l =0 being
l
centre of the ring) in volts is A. + 2
B. – 1
C. – 2
D. zero
10. A parallel plate capacitor of area A, plate separation d and capacitance C is filled with three different dielectric materials having constants K 1, K 2 and K 3 as shows. If a single dielectric material is to be used to have the same capacitance C, in this capacitor then its dielectric constant K is given by: A/2
A/2
A.
1 1 1 1 K K1 K2 2K 3
C.
K1 K 2 1 2K 3 K1 K 2 K
K2
K1
B.
1 1 1 K K1 K 2 2K3
d K3
d/2
D.
K1 K 3 K 2 K3 1 K1 K 3 K 2 K 3 K
A
Where A = area of plates
11. A uniform electric field pointing in positive x-direction exists in a region. Let A be the origin, B be the point on the axis at x =1 cm and C be the point on the y-axis at y =1 cm. Then the potentials at the point A, B and C satisfy: A. V A VB
B. VA VB
D. VA VC
C. V A VC
12. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to +q1 q2 -q1
A. q 2
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Electrostatics 5
13. Two fixed, equal, positive charges, each of magnitude 5 10 coulomb are located at points A and B separated by a distance of 6 m. An equal and opposite charge moves towards them along the line COD, the perpendicular bisector of the line AB. The moving charge, when it reaches the point C at a distance of 4 m from O, has a kinetic energy of 4 joule. Then, the distance of the farthest point D which the negati ve charge will reach before returning towards C
A +q O
D
A. 4.48 m C. 8.81
-q C
B. 2.24 m D. None
B 14. A parallel plate capacitor is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, the time rate of change of electrostatic energy of a capacitor is proportional to: A. x 2
C. x 1
B. x
D. x 2
15. In the circuit as shown in the figure, the effective capacitance between A and B is
4F
A. 8F
B. 4F
C. 2F
D. 3F
A
2F
2F
4F 4F
B 16. A cube of metal is given a positive charge Q. For the above system, which of the following statements is true? A. Electric potential at the surface of the cube is zero B. Electric potential within the cube is zero C. Electric field is normal to the surface of the cube D. Electric field varies within the cube 17. A charge (- q) and another charge (+Q) are kept two points A and B respectively. Keeping the charge (+Q) fixed at B, the charge (- q) at A is moved to another point C such that ABC forms an equilateral triangle of side l . The net work done in moving the charge (- q) is A.
1 Qq 4 0 l
B.
1 Qq 4 0 l 2
C.
1 Qql 4 0
D. zero
18. Two conductors of radius R1 and R2 capacitances C1 and C 2 have charges q1 and q 2 respectively when they are joined together by a conducting wire, charge redistributes in these conductors q '1 and q ' 2 respectively, then
q '1 = A.
R2 (q1 q 2 ) R1 R2
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R1 C2 (q1 q 2 ) C. (q1 q 2 ) R1 R2 C1 C 2
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Electrostatics 19. An isolated conductor initially free from charge is charged by repeated contacts with a plate which after each contact is replenished to a charge Q. If q is charge on the conductor after first operation, then maximum charge which can be given to the conductor in this way is A. Q
B. Q + q
C. zero
D.
Qq Qq
20. A capacitor is given a charge q. The distance between the plates of the capacitors is d. One of the plates is fixed and other plate is moved from the other till the distance between them 2d. Then work done by the external force is A. zero
B.
q2d A 0
C.
q2d 2 A 0
D.
2q 2 d A 0
21. A point charge q is placed on the top of a cone of semi vertex angle . The electric flux through the base of the cone A. zero
B.
q 2 0
C.
q
0
D.
q(1 cos ) 2 0
22. Incorrect statement is A. Electric field is in the direction in which the potential steepest. B. Magnitude of Electric field is given by the change in the magnitude of potential per unit displacement normal to the equipotential surface at the point. C. A charge is placed inside a cavity in conductor, then electric field is zero in cavity. D. Electric field at the surface of a charged conductor is normal to the surface and does not depend upon surface charge density
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Electrostatics Advance Assignment 2 Section I: Q. No. 1 to 8 has only one correct answer 1. Consider a neutral conducting sphere. A positive point charge is placed outside the sphere. The net charge on the sphere is then A. Negative and distributed uniformly over the surface of the sphere B. Negative and appears only at the point on the sphere closest to the point charge C. Negative and distributed non-uniformly over the entire surface of the sphere D. zero 2. A thin spherical conducting shell of radius R has a charge q. Another charge Q is placed at the centre of the shell. The electrostatic potential at a point P at distance
R from the centre of the shell is: 2
A.
(q Q) R 4 0 2
B.
2Q 4 0 R
C.
2Q 2q 4 0 R 4 0 R
D.
2Q q 4 0 R 4 0 R
3. A charge Q is distributed over two concentric hollow spheres of radii r and R(R>r) such that their surface densities are equal. Find the potential at the common centre k 4 0
1
A.
kQ Rr
B.
kQ( R r ) R2 r 2
C.
kQ r
D.
kQ R
4. Two identical thin rings, each of radiuses R, are coaxially placed a distance R apart. If Q1 and Q2 are respectively the charges uniformly spread on the two rings, the work done in moving a charge q from the centre of one ring to that of the other is:
2 4 R D. q(Q / Q ) 2 1 2 4 R B. q(Q1 Q2 ) 2 1 /
A. zero C. q 2 (Q1 Q2 ) / 4 0 R
1
2
0
0
5. Consider the situation shown in the figure. The capacitor A has a charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after the switch is closed is: q
+
-
A. zero
B. q/2
+
-
C.q
D. 2q
+
-
+
-
+
S
A
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Electrostatics 6. Two identical capacitors have the same capacitance C. One of them is charged to potential V1 and the other to V2 . The negative ends are also connected, the decrease in energy of the combined system is: A.
1 C V12 V22 4
B.
1 C V12 V22 4
C.
1 C V1 V2 4
2
D.
1 C V1 V2 4
2
7. The capacitance of parallel plate capacitor with plate area A and separation d is C. The space between the plates is filled with two wedges of dielectric constants K 1 and K 2 , respectively. The capacitance of resulting capacitor is: A
K1
K2
K1 K 2 A 0 K log 2 d ( K1 K 2 ) K1 d K1 K 2 A 0 K C. log 2 d ( K1 K 2 ) K1
K1 K 2 A 0 K log 1 d ( K1 K 2 ) K2 K1 K 2 A 0 K D. log 1 d ( K1 K 2 ) K2
A.
B.
a a 2 2 work done by the electric field when another positive charge is moved from (a,0,0) to (0, a,0) is positive
8. Positive and negative point charges of equal magnitude are kept at 0,0, and 0,0, , respectively. The
A. positive C. zero
B. negative D. depends upon the path connecting the initial points
Section II: Q. No. 9 to 12 have one or more correct answer 9. A conducting sphere S1 and radius r is attached to an insulating handle. Another conducting sphere S 2 of radius R is mounted on an insulating stand. S 2 is initially uncharged. S1 is given a charge Q. brought into contact with S 2 and removed. S1 is recharged such that the charge on it is again Q; and it again brought into contact with S 2 and removed. This procedure repeated n times
QR R 1 A. The electrostatic energy of S 2 after n such contacts with S1 is 4 0 R r R r 1
QR R 1 B. The electrostatic energy of S 2 after n such contacts with S1 is 8 0 R r R r 1
C. The limiting value of this energy as n is
Q2R 4 0 r 2
Q2R D. The limiting value of this energy as n is 8 0 r 2
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2
2
Electrostatics 10. A parallel-plate capacitor is connected to a battery. A metal sheet of negligible thickness is placed between the plates. The sheet remains parallel to the plates of the capacitor. A. The battery will supply more charge B. The capacitance will increase C. The potential difference between the plates will increase D. Equal and opposite charges will appear on the two faces of the metal plate. 11. The electric field in a region is directed outward and is proportional to the distance r from the origin. Taking the electric potential at the origin to be zero, A. it is uniform in the region B. it is proportional to r C. it is proportional to r 2 D. it increases as one goes away from the origin 12. An electric dipole is placed at the centre of a sphere. Mark the correct options: A. The flux of the electric field through the sphere is zero B. The electric field is zero at every point of the sphere C. The electric field is not zero anywhere on the sphere D. The electric field is zero on a circle on the sphere Section III: Passage for Q. 13 to Q. 18 Passage for from Q. 13 to Q. 15 The nuclear charge (Ze) is non-uniformly. Distributed within a nucleus of radius R. The charge density (r ) {charge per unit volume} is dependent only on the radial distance r from the centre of the nucleus as shown in figure. The electric field is along the radial direction
(r ) d
r a
R
13. The electric field at r = R is
`
A. independent of a
B. directly proportional to a
C. directly proportional to a 2
D. inversely proportional to a
14. For a = 0, the value of d (maximum value of as shown in the figure) is A.
3Ze 4R 3
B.
3Ze R 3
C.
4 Ze 3R 3
D.
Ze 3R 3
15. The electric field within the nucleus is generally observed to be linearly dependent on r. This implies A. a = 0
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C. a = R
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D. a
2R 3
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Electrostatics Passage for Q 16 to 18 Two points A and B are 2 cm apart and a uniform electric field E acts along the straight line AB directed from A to B with E = 200N/C. A particle of charge 1 10 6 C is taken from A to B along AB.
16. The force on the charge B. 1 10 6 N
A. Zero
C. 2 10 6 N
D. 2 10 4 N
C. 3 volt
D. 4 volt
C. 4 10 6 J
D. 1 10 6 J
17. The potential difference V A VB A. Zero
B. 2 volt
18. The work done on the charge by electric field B. 2 10 6 J
A. Zero
Section IV: Q. No. 13 to 14 are statement – 1 (Assertion) and Statement – 2 (Reason) type questions and carries 3 marks each for correct answer and –1 mark for each wrong answer. The following six problems consists two statements. You have to examine these two statements carefully and decide if the Assertion and Reason are individually true and if so, whether the Reason is the correct explanation of the Assertion. Select your answer to these problems using the code given below and mark answer accordingly: A. Statement -1 is True, Statement – 2 is True; Statement -2 is correct explanation for statement 1. B. Statement-1 is True, Statement –2 is True; Statement-2 is not correct explanation for statement-1 C. Statement -1 is True, Statement – 2 is False D. Statement -1 is False, Statement – 2 is True 19. STATEMENT-1 For practical purposes, the earth is used as a reference at zero potential in electrical circuits. STATEMENT-2 The electrical potential of a sphere of radius R with charge Q uniformly distributed on the surface is given by
Q . 4 0 R
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Electrostatics 20. Statement 1: A charge q is placed at the centre of a metallic shell as shown in figure. Electric field at point P on the shell due to charge q is zero. Statement 2: Net electric field in a conductor under electrostatic conditions is zero.
P
q
A. A
B. B
C. C
D. D
E. E
Section V: Match the column 21. Column I
Column II
A. Equipotential surface is always
p. Vector
B. Electric potential gradient is
q. normal to electric field
C. In series combination of capacitors, what is same on each capacitors is
r. potential difference
D. In parallel combination of capacitors, what is same for each capacitors is
s. charge
22. Two spherical shells are shown in figure, suppose r is the distance of a point from their common centre. Then,
q2
q1
R1 = radius of inner shell
R2
Column I
Column II
A. Electric field for r R1
p. is constant for q2 and vary for q1
B. Electric potential for r R1
q. is zero for q2 and vary for q1
C. Electric potential for R1 r R2
r. is constant
D. Electric field for R1 r R2
s. zero
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Electrostatics Advance Assignment 3 Section I: Q. No. 1 to 8 has only one correct answer 1. A long, hollow conducting cylinder is kept coaxially inside another long, hollow conducting cylinder of larger radius. Both the cylinder is initially electrically neutral. A. A potential difference appears between the two cylinders when a charge density is given to the inner cylinder B. A potential difference appears between the two cylinders when a charge density is given to the outer cylinder C. No potential difference between the two cylinders when a uniform line charge is kept along the axis of the cylinders D. No potential difference appears between the two cylinders when same charge density is given to both the cylinders. 2. Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at O is double the electric field when only one positive charge of same magnitude is placed at R. Which of the following arrangements of charge is possible for, P, Q, R, S, T and U respectively? P
Q
O
U
T
A.
R
S
,,,,,
B. ,,,,,
C. ,,,,,
D. ,,,,,
3. A spherical portion has been removed from solid sphere having a charge uniformly distributed in its volume as shown in the figure. The electric field inside the emptied space is
A. zero everywhere C. non-uniform
B. non-zero and uniform D. Zero at only its centre
B
4. The electrostatic potential V at any point (x,y,z) in space is given by V 4x 2 . A. They y – and z- components of the electrostatic field at any point are zero
B. The x-component at any point is given by 8 x i
C. The x-component at a point (1, 0, 2) is 8 i D. They y- and z-components of the field are constant in magnitude.
5. The magnitude of electric field E in the annular region of a charged cylindrical capacitor: A. is same through
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Electrostatics B. is higher near the outer cylinder than near the inner cylinder C. varies as 1/r where r is the distance from the axis D. varies as 1 / r 2 where r is the distance from the axis 6. Two isolated metallic spheres of radii R and 2R are charged such that both of these have same charge density . The spheres are located far away from each other, and connected by a thin conducting wire. Then, new charge density on the bigger sphere A.
B.
6
7 6
C.
5 6
D. None
7. Two uniformly charged plane sheets S1 and S 2 having charge densities 1 and 2 ( 1 2 ) are placed at a distance d parallel to each other. A charge q 0 is moved along a line of length a (a < d) at an angle 45 0 with the normal to S1 . Then the work done by an electric field A.
q0 ( 1 2 )a
B.
q0 ( 1 2 )a
2 2
C.
q0 ( 1 2 )a
2 2
D. None
2
8. Two equal point charges are fixed at x a and x a on the x-axis. Another point charge Q is placed at the origin. The change in the electrical potential energy of Q, when it is displaced by a small distance x along the x -axis, is approximately proportional to: B. x 2
A. x
C. x 3
D. 1/x
Section II: Q. No. 9 to 12 have one or more correct answer
Q 4 R 0 0 9. A spherical symmetric charge system is centered at origin. Given Electric potential V Q 4 0 r V
R0
r
A. within r 2R0 total enclosed net charge is Q
B. Electric field is discontinued at r R0
C. Change is only present at r R0
D. Electrostatic energy is zero for r R0
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r R0 r R0
Electrostatics 10. A charge + Q is fixed at the origin of the coordinate system while a small electric dipole of dipole moment
P pointing away from the charge along the x axis is set free from a point far away from the origin. 1 pQ A. The kinetic energy of the dipole when it reaches to a point (d, 0) is 4 0 d 2 B. The kinetic energy of the dipole when it reaches to a point (d, 0) is C. The force on the charge + Q at this moment is
1
2 pQ 4 0 d 2
2 pQ 4 0 d 3
D. The force on the charge + Q at this moment is
1
1
pQ 4 0 d 3
11. Four point charges +8mC, - 1 mC, - 1mC, and + 8 mC are fixed at the points
27 3 3 m, m, m and 2 2 2
27 m respectively on the y axis. A particle of mass 6 10 4 kg and charge 0.1C moves along the – x direction. 2 Its speed at x is v 0 . (Assume that space is gravity free) A. The least value of v 0 for which the particle cross the origin is 3 m/sec B. The least value of v 0 for which the particle cross the origin is 4 m/sec C. The kinetic energy of the particle at the origin is 3 10 4 J D. The kinetic energy of the particle at the origin is 5.3 10 4 J
12. The separation between the plates of a charged parallel-plate capacitor is increased. Which of the following quantities will change? A. charge on the capacitor
B. potential difference across the capacitor
C. energy of the capacitor
D. energy density between the plates
Section III: Passage for Q. 13 to Q. 18 Passage for Q 13 to 15 A, B, C, D are four thin, similar metallic parallel plates, equally separated by distance d, and connected to a cell of p.d.(V), as shown in figure.
A
C D
B 1
2
3
V
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Electrostatics 13. The potential of plate D A. zero
B. V
C. V/2
D. None
14. If B and C be connected by a wire, then what will be the potential of plate D A. zero
B. V
C. V/2
D. None
15. How will the electric field will change in the spacing between the plates B and C A. zero
B. Increase
C. decrease
D. None
Passage for Q 16 to 18 In an isolated system (neither connected to the terminal of a battery nor to any other source of charge e.g. earth) net charge remains constant. From the two terminals of a battery or from two plates of capacitor equal and opposite charges enter or leave. Two capacitors of capacity 6F and 3F are charged to 100 V and 50 Volt separately and connected as shown in figure. Now all the three switches S1, S 2 and S 3 are closed
6F 1 2
S2
3
S1
100V
4 50V
3F S3
200V 16. Which plate (s) form an isolated system: A. Plate 1 and plate 4 separately
B. Plate 2 and plate 3 separately
C. Plates 1 and 4 jointly
D. Plates 2 and 3 jointly
17. Charge on both the capacitor in steady state will be (on 6F first): A. 400C , 400C
B. 700C , 250C
C. 800C , 350C D. 300C , 450C
18. suppose q1 , q 2 and q 3 be the magnitude of charges flow through switches S1 , S 2 and S 3 after they are closed. Then: A. q1 q3 and q 2 0 C. q1 q3 2q2
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Electrostatics Section IV: Q. No. 13 to 14 are statement – 1 (Assertion) and Statement – 2 (Reason) type questions and carries 3 marks each for correct answer and –1 mark for each wrong answer. The following six problems consists two statements. You have to examine these two statements carefully and decide if the Assertion and Reason are individually true and if so, whether the PReason is the correct explanation of the Assertion. Select your answer to these problems using the code given below and mark answer accordingly: A. Statement -1 is True, Statement – 2 is True; Statement -2 is correct explanation for statement 1. B. Statement-1 is True, Statement –2 is True; Statement-2 is not correct explanation for statement-1 C. Statement -1 is True, Statement – 2 is False D. Statement -1 is False, Statement – 2 is True 19. Statement 1: The dielectric constant of a conductor is infinite. Statement 2: Dielectric constant depends on the availability of free electrons in the substance. A.
B.
C.
D.P
20. Statement 1: A parallel plate capacitor is connected across a battery through a key. A dielectric slab of constant k is introduced between the plates. The energy stored in the capacitor becomes k times. Statement 2: The surface density of charge on the plates remains uncharged. A.
B.
C.
Section V: Match the column 21. Column I
Column II
A. Electric field is a vector whose dimensions are
p. ML3T 3 A1
B. Electric flux is a scalar whose dimensions are
q. ML2T 3 A1
C. Dimension of electric dipole moment
r. MLT 3 A1
D. Dimension of Electric potential is
s. M 0 LTA
D.
22. In the figure shown P is a point on the surface of an imaginary sphere, Match the following
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Electrostatics q2 q1
P
Column I
Column II
A. Electric field at point P
p. due to q1 only
B. Electric flux through a small area at P
q. due to q2 only
C. Electric flux through whole system
r. due to both charges
Advance Assignment 1 1. D 15. B
2. B 16. C
3. D 17. D
4. A 5. B 18. B, D
6. B 19. D
7. B 20. C
8. C 21. D
9. A 22. D
10. D
11. B
12. C
13. A
14. A
6. C 19. B
7. B 20. D
8. C 9. B, D 10. D 11. C 12. A, C 13. A 21. A – q, B – p, C – s, D – r 22. A – s, B – r, C – p, D – q
Advance Assignment2 1. D 14. B
2. D 15. C
3. B 16. D
4. B 17. D
5. A 18. C
Advance Assignment 3 1. A 2. D 3. B 4. A,B,C 5. C 6. C 11. A, C 12. B,C 13. A 14. A 15. A 21. A – r, B – p, C – s, D – q 22. A – r , B – r, C - p
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8. D 17.B
9. A,B,D 18. D 19. B
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