# electrostatics.pdf

November 15, 2017 | Author: AshutoshYelgulwar | Category: Electric Charge, Electron, Proton, Mass, Matter

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ELECTROSTATICS

SESSION 1 AIM  To introduce the concept of charge and properties of charge  Compare charge with the mass  To introduce Coulomb’s Law  To introduce Principle of Superposition ELECTROSTATICS It is the branch of physics where we study the behavior of charges at rest. Charge: What is charge? Charge is an inherent property of matter. All the matter in the universe is composed of atoms (combined as molecules). These atoms have protons and neutral neutrons forming nucleus surrounded by electrons orbiting around the nucleus. Other than mass protons and electrons have an additional property known as charge which results in a force between these particles like the gravitational force arising due to mass. This force is called electrical force. There are two kinds of charges in these particles. One is that possessed by electrons and the other possessed by protons. Neutrons don’t have such property known as charge. Benjamin Franklin observed that these two kinds of charges can cancel the effect of each other if they are put together, just like the negative and positive numbers ahen added. Based on this observation he introduced the concept of positive and negative charges. It is purely a matter of chance that proton is

assigned with positive sign and electrons were given the negative sign and there is no logic at all behind this. For neutral atoms (Matter), number of electrons is always equal to number of protons. Therefore net charge is zero. charge of proton = +1.6 × 10 ; charge of electron = −1.6 × 10 ; Since protons are confined to nucleus, and due to strong nuclear forces these protons can not break apart inspite of proton to proton repulsion. the charge is always produced by transfer of electrons only. ‘-ve’ charge ‘+ve’ charge 1. Gain of electrons 1. Loss of electrons 2. Gain of matter. So 2. Loss of matter. So mass of negatively mass of positively charged body is charged body is increased decreased Properties of Charges and its comparison with mass: 1] Charge is called electrical property and mass gravitational property of matter. 2] There are two kinds of charges namely positive and negative. Mass is of one type positive. 3] Like charges repel each other and unlike charges attract each other. In case of mass it is always attraction. 4] Charge is additive by nature that means in a body total charge can be obtained by calculating algebraic sum of all charges, considering their positive and negative sign. In case of mass there is nothing like that to consider as mass is always one type and is to be added always.

5]

Charge is quantized but quantization of mass is not yet established. What is quantization? There is a minimum value of charge that can exist. And this minimum amount is equal to the charge of an electron. Electron is not a particle which can be further cut into pieces. Charges appearing on an object is always due to transfer of electrons. So charge in a body can always exist in integral multiple of charge on an electron, which can be positive or negative depending on whether the electrons are lost or gained by the object. So any charge can be expressed as q = ne where n is a whole number. Note: Mass of each electron is me 9.1x10_31kg. So whenever n electrons are lost by the body then charge acquired will be q=+ne and mass decreases by nme. Note: Do you know money is quantized. Suppose you have a 100 Rs. note and you want to distribute it among four persons. You can easily tear off the note into four equal parts. Are you done distributing? you have literally destroyed 100 Rs note and no one will gain anything. Other option seems wiser. You can get the change and then distribute it. Did you understand? So quantization means fixed denomination. Luckily for charge we have only one value (not like money 2 Rs. 5 Rs, 100 Rs, 500 Rs, 1000 Rs etc.) and that value is the charge of electron (e = 1.6 10_19C) So any charge can be expressed as q = ne where n is a whole number.

Therefore “Quantization of charge” can be explained as “The charge, which can be transferred or may exist freely in nature, will always be an integral multiple of the charge of electron.” Note: The charges on the polar ends of dipole (covalent molecules) may be less than that of electron, leading to % ionic character of molecule but whenever bond is broken, charge shared by fractions is an integral multiple of the electronic charge. 6] Charge is conserved but mass is not. As we know that as per Einstein’s equation E = mc2 mass and energy are interconvertible. As per the equation if we annihilate the mass m then a total of mc2 energy is generated. On the other hand if

7]

we talk about the conservation of charge then let me tell you that the concept is valid for an isolated system. Net charge of an isolated system remains conserved. But there is nothing like that for net negative or net positive charge. For example in case of neutron annihilation process one positron and one beta particle is generated. Hence the system was neutral before that and neutral after that but one -ve charge is created and one positive charge is created. So conservation of charge is only for net charge but not for individual positive or negative charge or number of charged particles. Charge is invariant but mass is not. Invariance means when body moves then there is no change in magnitude of charge. But mass changes with velocity. , this equation shows the variation of mass with velocity of the object. Here v is the speed of the body and c is the speed of light. mo is the rest mass of the body.

Frictional Electricity: Most basic method of creating charge is by friction. When two suitable object are rubbed together against each other than our work done results in the form of heat and this energy is absorbed by the electrons and electrons are transferred from one object to another and it causes the charging by equal amount of opposite nature in these two objects. One in which electrons are loosely bounded losses the electrons and attains positive charge and the other gains electrons and gets negatively charged. And that is how the charged bodies are created in nature. Examples: Ex1: Rub a piece of ebonite (very hard, black rubber) across a piece of animal fur. The fur does not hold on to its electrons as strongly as the ebonite. At least some of the electrons will be ripped off of the fur and stay on the ebonite. Now the fur has a slightly positive charge (it lost some electrons) and the ebonite is slightly negative (it gained some electrons).The net charge is still zero between the two… remember the conservation of charge. Ex2: Rub a glass rod with a piece of silk. This is the same sort of situation as the one above. In this case the silk holds onto the electrons more strongly than the glass. Electrons are ripped off of the glass and go on to the silk. The glass is now positive and the silk is negative. But the question is how to remember that which object is going to loose the electron and which one is going to gain. For that you can remember the following electrostatic series.

Charging by Induction: In the induction process, a charged object is brought near but not touched to a neutral conducting object. The presence of a charged object near a neutral conductor will force (or induce) electrons within the conductor to move. The movement of electrons leaves an imbalance of charge on opposite sides of the neutral conductor. While the overall object is neutral (i.e., has the same number of electrons as protons), there is an excess of positive charge on one side of the object and an excess of negative charge on the opposite side of the object. Once the charge has been separated within the object, a ground is brought near and touched to one of the sides. The touching of the ground to the object permits a flow of electrons between the object and the ground. The flow of electrons results in a permanent charge being left upon the object. When an object is charged by induction, the charge received by the object is opposite the charge of the object which was used to charge it.

Coulomb’s law: The force of interaction between two charges ( , ) separated by a finite distance(r) in space is directly ) and proportional to the product of charges ( ∝ inversely proportional to square of their separation. ∝

(Inverse square law); Combining the two

=

Where ‘k’ is the electrostatic coefficient. = 9 × 10

;=

; Where

= 8.854 × 10

/

is

called “Permittivity of free space” Note:  for like charges > 0; repulsion  for unlike charge < 0; attraction  Electrostatic force is a kind of central force i.e. independent of presence of other charges except the pair under consideration. Principle of Superposition: Suppose a charge is under influence of forces ⃗ , ⃗ , ⃗ . . . . . . ⃗ Net force will be → ⃗

= ⃗ + ⃗

CLASS EXERCISE 1] Which of the following charge is not possible. If it is possible, then how to develop this charge. a) 10 b) 10 Sol: a) q = ne n= =

; not an integer, so not possible,

. ×

b) = = . × ( ‘+ ve charge)

; = 6.25 × 10

electrons must be removed

2]

Find the minimum electrostatic force acting between two charges at a separation of 1m . Sol: Here force can take any value but minimum possible amount of charge is e = 1.6 × 10 So Fmin = 3]

=

×

×

. ×

= 2.304 × 10

.

Three identical charges q each are kept at vertices of an equilateral triangle of side a. Find the force on any charge by the other two. How the system can be held in equilibrium. F FA

FB

C a 60

Sol:

600 0

a 60 0

a

B

Since

=

A

⃗ = ⃗ + ⃗ (Net force) =

; FNet = 2FA cos (60/2)

(Law of parallelogram for two forces of equal magnitude) F = 2 .

.

=

To maintain the equilibrium, a charge of opposite nature must be placed at the centroid. Let the charge be – F q

a

F'

3

 q'

q √

q for equilibrium; –

=

= −

⃗ = – ⃗ ; ;

=

=

So charge at centroid must be

SESSION – 2 AIM  To introduce the Vector form of Coulomb’s Law.  To solve problems based on the concepts learnt in the previous session regarding principle of superposition but of higher difficulty. (Including Continuous charge distribution) THEORY Vector form of Coulomb’s Law: q2

y q1 r1

F2 / 1

r

F1 / 2

r2 x

z

/

= − ⃗ / (Force on q1 by q2) = – (Force on q2 by q1) = ⃗/ = . ̂; ⃗ = ⃗ – ⃗ ; ⃗

/

=

.

⃗ – ⃗

⃗ – ⃗

We can use this equation only when position coordinate of charges are given. In general we consider direction and resolve the forces suitably so that we can find an easier solution.

CLASS EXERCISE 1] The distribution of charges in space are at +q[a,0,0], –q[0,a,0], +q[0,0,a] and +q[a,a,a]. Find net force on +q[a,a,a] Let ⃗ be force between +q[a,0,0] and +q[a,a,a]; be force between –q[0,a,0] and +q[a,a,a] be force between +q[0,0,a] and +q[a,a,a]; ⃗ = ⃗ + ⃗ + ⃗ = = 2]

+

̂+

2 ̂ =

+ ( ̂ + ̂) ;

̂+

̂

Find the force between a point charge q and a linear charge of length l and linear charge density for following cases. 

A)

q

.

a 

q

a

l

B)

l

Sol: (A) Take point charge at origin and consider an element of length dx on linear charge at distance x from origin. l q a

(a,0) dx

(l+a,0)

x

Charge on element dq = dx force on point charge dF = So net force =

(Linear charge lies between x = a to x = l + a) = =

(

− )

; if a>>> l (small linear charge)

=

=

=

Hence again we get same simpler result. (B) Consider linear charge kept on x-axis with one end at origin. Take one element of length dx at distance x from origin. dF

. 

dFx

dFy

r a

(0,0)

(,0) l

dx l x

From the triangle shown above; x = a tan ; dx = a sec2 r = a sec So dq = dx = a sec2 d There fore dFy = dFcos ; ∴

=

=

[

=

.

.

]

;

0

;

;

= =

Similarly =∫ =

1 –

So ⃗ = −

−1 √ [− cos ] ; 0

= √

̂ +

̂

d ;

3]

Find the force between a point charge q kept at centre of an arc of radius R and angle 2 with linear charge density .

Consider a line of symmetry (OA) and two symmetrical elements at ∅ and ∅ is the angle subtended at the centre ‘C’ by the symmetrical elements. Charge of these elements dq = = ∅ Force between element and point charge q ∅

=

=

From figure you can assess that forces ⊥ to line of symmetry OA are getting cancelled due to symmetry. The forces parallel to OA are going to produce net force Hence = ∫ 2 ∅ =∫

∅ ∅

=

SESSION – 3 AIM  To introduce the concept of electric field  Lines of Electric field  To introduce the concept of applying superposition principle to find the resultant electric field. THEORY 1] Electric field: The region in space where an object experiences (electrostatic) force by the virtue of its charge is called Electric field. Quantitatively, Electric field is the force experienced per unit charge. ⃗ =

where (q) charge is placed at position where the field has to the determined. Test Charge: The point charge which is placed at the position where Electric field is to be determined. Source charge (s): The charge which is the cause behind development of field. The charge (s) other than test charge (q) are called source charge. Note: In all the examples in last sessions (Electrostatics session 1 & 2), if we divide the force by any point charge, we will obtain the electric field at the position of point charge. Since field is position specific phenomenon, test charge must be a point charge.

A) Electric field of point charge (Q). Assume a point test charge q at a distance r from source charge Q. The force of mutual interaction between the two is given as = position of

; So, field at position of ⟹

=

=

and at

=

Note: Since source charge here is also a point charge, Electric field can be determined at the position of charge Q by ⃗ =

I)

; ⃗ and

⃗ here must have the same direction if q is positive. i.e. Electrostatic force on a positive charge is along the direction of electric field but the electrostatic force on a negatively charged body is opposite to electric field. Now, we can determine the direction and magnitude of Electric field for all the cases discussed in session 1 and 2 of Electrostatics. The electric field determination is carried out by two methods (I) Coulombian Approach (II) Gaussian Approach Coulombian Approach: Here we consider the field of charge element and integrate it over length, surface or volume. This approach is a little bit complicated but any type of charge distribution can be dealt with it successfully provided system is at least geometrically defined and is not any irregular shape. The examples here will give you an idea to use this technique and to reduce 2-D and 3-D systems to simple charge distributions in 1-D integration.

Note: Some of examples of session - 2 are already based on Colombian approach. Other examples will be discussed in session - 4 Lines of Electric Field: The imaginary lines that give the direction of electric field at any point in the space and also tell us about its relative strength are called electric field lines. The tangent to the electric field line at any point gives us the direction of electric field. When the lines of electric field are closer the field strength is higher. 1] Lines of electric field are always directed outwards from isolated +ve charges, normal to surface of conductor and inwards for isolated negative charge.

Point or spherical

 + + + + + + + + + +

+ + + + +

Plane Surface

Side View ++ ++ ++ + + ++ ++ ++

Top View

linear or cylindrical

2]

For a pair of opposite charges, lines of electric field originate at positive charge and terminate at negative charge. Keep in mind, no of lines must correspond to amount of charge. + 3q

3] 4] 5]

6]

q

From above figure you can see lines of electric field tend to contract lengthwise and expand sidewise. The tangent at any point will represent direction of electric field. Two lines of electric field will never intersect because that will represent two different directions of field at a place, which is impossible for a vector quantity Density of lines of electric field correspond to the magnitude of electric field P

7]

Q

R

EP > EQ > ER Lines of electric field do not exist within a conductor or conducting cavity which do not contain electric charge and lines of electric field always are normal to the conducting surface.

Uniform Electric Field

Conductor in uniform electric field

You can easily notice that field within conductor in such case is zero. Induction of charge will take place but net charge induced will be zero.

8]

Dielectric: The materials that try to resist the induction (usually non conductor, covalent molecules) and tend to decrease the electric field are dielectrics. E

E

Uniform Electric Field

Dielectric in Uniform Electric Field

Since the electric field is lesser in dielectric, we use an additional term here and replace by = = ; K = Dielectric constant or relative permittivity; K = 1 for air; K = ∞ for conductor. This modification can be used to determine field, force and potential in dielectric medium. Special Cases: 1] Field due to two like charges: Null point x1 x2

+

E1 E2 P

Q1

Q2

d

At null point ⃗ = 0; Force experienced by any charge ⃗ = 0 You can notice here that no line passes through a point (P in figure), which is known as null point charges, it lies on the line joining the charges, between the charges. At this point electric field is zero. The point is closer to smaller charge in magnitude. ⃗ = − ⃗ at null point and ⃗ = ⃗ ; = =

and x1 + x2 = d;

⇒ x1 =

and x2 =

2]

Field due to two unlike charges; Null point

Again point P is a null point lying on the line passing through , outside the charges, near smaller charge, at point P, ⃗ = − ⃗ and ⃗ = ⃗ = and ⇒

=

√ √

=

and x1 – x2 = d; ⇒

Electric field due to three like charges

+

3]

;

P

+

Point P is null point At null point ⃗ = 0; ⃗ = 0

=

√ √

CLASS EXERCISE 1] Suppose all surrounding charges are fixed to their position and when a test charge +q is placed at point P then it is observed that the electric field at point P is E. Now if the magnitude of test charge is doubled then find the electric field at point P. 2] Find the electric field at corner P. +q 60

q

3]

o

60o (P) +q

60o

Find the electric field at corner P. +2 2q

a

+q

a

a (P)

-q

4]

a

+q

Find the electric field at corner P. +q a

a a

-2q a

P a

+q

5]

Find the electric field due to line charge.  +

+

+ + + + + + + +

l

Find electric field at point P...

P a

= charge per unit length

SESSION – 4 AIM  To introduce problem solving approach for finding electric field due to charged ring.  To introduce problem solving approach of finding the electric field due to charged disc.  To introduce the expression of electric field due to a hemispherical shell.  To deduce field due to a linear conductor. 1]

Electric Field Due to a Uniformly Charged Ring: A ring of radius R is uniformly charged by a charge Q. Electric field due to this ring can be discussed in two cases. Case-I: Electric field at centre: Due to symmetry electric field at centre of ring is always zero for uniform distribution. Note: For non uniform charge distribution, solve as field due to an arc at centre of arc. Case-2: Electric field on the axis of ring: Again we consider two diametrically opposite elements subtending an angle at centre or length each.

So and

=

= =

= ; =√

= +

= =

(

)

Note: For non IIT students derivations are not required keep result in mind. You can visualize easily that component of electric field ⊥ to axis (dE sin ) are cancelled due to symmetry and it is the only components || to axis dE cos which activity contributing for net electric field. Hence = ∫ 2 ⇒ 2]

=

;

(

; =

)

.

) /

(

Electric field at the axis of a uniformly charged disc: Electric field due to a disc is to be determined at its axis. We can split the disc into concentric rings because we know the electric field for ring. Such ring will have a charge dQ, so field is also

=

(

) r

dE

R

x

The ring considered here is of radius r and thickness dr so dQ = dA =

.2

So = ∫

.

; =

.

let r = x tan r = 0 ⇒ dr = x sec2 d x2 + r2 = x2 sec2

; ) /

(

=0 r=R⇒

= cos–1

So =

.

=

=

[−

2

= =

]

1− 1−

;

√ + 0 if =

if R >>> x in finite sheet = 3]

Electric field at centre of a uniformly charged hemispherical shell: Consider a uniformly charge spherical shell of surface charge density. Again we can split the shell into thin rings.

Let us take a ring of radius r at distance x from centre of thickness R d r = R sin from figure; dA = 2 r . Rd ; dA = 2 R2 sin d ; dQ = dA = 2 sin ; x = R cos ; So

=

(

(

) ) /

=

;∫

= If

/2 . [− cos 2 ] = 0 = =

/

= ∫

[−(−1) + 1] =

; =

Note: If we cut this hemisphere in two equal parts the field due to each will he

.

E  4 0 E

= 4]

=

=

;

√2 =

,

=

Electric field due to linear uniformly charged body. Case-1 Field due to linear charge on its Axis. Consider a linear charge of length l and linear charge density .

(- l/2)

The point P lies on the axis of linear charge, at a distance d from centre. To calculate electric field at point P assume a small element at length d at (x, 0). Its charge is dQ = dx Its distance from the point P = ( − ) So Electric field = = for net Electric ( ) ( ) field integrate both sides =

/ \ (

)

;=

=

(

/ )

;

Case-2: Electric field at perpendicular bisector of linear charge: Consider a line charge of length and linear charge density . The point P lies at distance on its perpendicular bisector. dEy dE P

dE

dEx

dEx



d

θ

r dx

( x, 0) (-//2, l 0)

dx

x

(+//2, l 0)

To find the electric field at point P consider a small element at distance ‘x’ from mid point of length dx. Charge on the element dQ = dx but x = d tan from figure dx = d sec2 d r = d sec So dQ = d sec2

d hence =

;=

;=

you can see component of Electric field parallel to charge (dEx) will be cancelled due to symmetric element on (–x, 0) hence net field is contributed by component perpendicular to charge (dEy). dEy = dE cos hence net field = ∫ cos + ] = ∫ ;= [ = − / / but = ;⇒ = . if >>> /

/

charge =

.

;

=

(When l >>d)

Note for semi infinite linear charge

Infinite

450

= =

= if d >>>

=

why? Half the charge. and

; for l ∅ ⟹ is - ve; iv) If ∅ = ∅ ⟹ = 0; v) Q= 0 does not mean no charge. Possibilities are there that net positive charge is equal to net negative charge enclosed within the surface. Special use of solid angle for Electric flux Theory: Suppose a point charge Q is placed before a circular hole of radius ‘a’ at a distance ‘d’ right before the centre.

Q

a

= ;

d

=

From the figure you can see semi vertex angle of cone formed in figure. We can convert it into solid angle and find the Electric flux passing through the hole. Think of a spherical surface of radius R. Cut this surface by a plane in such a way that area of intersection is in the form of a circular disc of radius a, at distance ‘d’ from centre. Rd 

R 

r =R sin 

 d a

The area of spherical portion will give the value of solid angle as ⇒ Ω = To calculate area, we split it into thin rings. One such ring subtend an angle ϕ at centre. Radius of this ring ⇒ = sin ϕ at centre. Thickness of this ring = ϕ So, Area of this ring = 2 . Rd ϕ = 2 R2 sin ϕ d ϕ; =∫ 2

sin ϕ dϕ

∅] = 2 [1 − ] 0 Therefore the solid angle formed from cone of angle = 2

=Ω=

= [ 1 −

=

[

]

= 2 (1 −

So flux through the hole ϕ = ∈ ∙

) = 2 [1 −

(4 = total solid angle)

],

ϕ=

1−

Let’s solve some problems... a

Q a

i)

L

Find the Electric flux a) Passing through the ends of cylinder b) Passing through the curved surface of cylinder a

d

ii)

Q

Charge Q1 is placed at the bottom of flower-Vase. Find the Electric flux a) Escaping from the vase. b) Passing through the surface of Vase. CLASS EXERCISE Electric field in a region is given as ⃗ = ̂ . Find charge enclosed within the surfaces x = 0, x = a, y = 0, y = a, z = 0, z = a. Sol: The planes described generate a cubical space with one vertex at origin. 1.

y

x z

at x = 0, ⃗ = 0; at x = a ; ⇒ ⃗ = ̂; Area at x = 0, ⃗ = ̂ ; Area at x = a,

2.

at x = 0 ⃗ = ̂; ∅ at x = a ⇒ = ̂. ̂ = E0 a4; ∅ = E0a4 Q= ∅= If field in a region is radially outward from origin, given as E = 1−

(for r < R) and E= 0 for r ≥ R at distance r from

origin find a) Flux passing through a spherical surface of radius r, center at origin. b) charge enclosed within radius r c) Value of r when charge begins to change the sign (Maximum charge) d) Charge density as a function of r. Sol: Area vector for spherical surface is also radially outward, Hence | | to electric field a) ∅ = ⃗ . ⃗ = b) =

1−

4

∅=4

=4

Applying Maxima and Minima. c)

=0⇒2 −

d)

=

− =

=0⇒ =

=

=4

=

3)

Find the total flux coming out of a cube if charge Q is placed a) at centre ⇒ ∅ b) at the edge ⇒ ∅ c) at the face ⇒ ∅ d) at the vertex⇒ ∅ ⇒∅

=

,∅

= .

= . = .

for all plane faces of any shape

= . ; Adjacent faces to the vertex

= 0, other three faces ∅

= . .

=

,

SESSION – 7 AND 8 AIM

       

Applying Gaussian Approach for determination of Electric field Point charge Charged conducting sphere / Shell Uniformly charged Spherical Volume Spherical Cavity Infinite Linear Charge Cylindrical Shell / Conducting Cylinder Uniformly charged Cylindrical Volume Cylindrical cavity

THEORY 1] For point charge: Assume a spherical Gaussian surface with charge at centre and passing through point of interest. Q

∮ ⃗. ⃗ = 2]

;⇒

∮ ⃗=

1; ⇒

r

=

For uniformly charged shell / charged conducting sphere: In both the cases net charge will come at outer surface only. So within the body charge will be zero. Assume a spherical Gaussian surface with centre at charge and passing through the point of interest.

Case 1: Inside r < R R r

∅ ∫ ⃗. ⃗ =

≠0 ℎ ℎ

; ⇒ = 0.

.

For out side r  R

∮ ⃗. ⃗ = ⇒ .4 3]

⃗ | | ⃗ =

;⇒

⃗ = =

⇒∮

=

For Uniformly charged sphere (Non conducting) charge density ( ) =

All the steps are like above but amount of charge has to considered properly. for inside r < R Qin = .

=

Therefore ∮ ⃗. ⃗ =

/ , . 4

=

; =

=

,

For outside r > R ⇒ ∮ ⃗ . ⃗ =

Qin = Q, . 4

=

, =

Note: For spherical cavity in above case apply superposition. ⃗ = ⃗ − ⃗

r d

Let density be ⃗

=

− ⃗

P

 r d

a

d

uniform.

; ⃗

= ; ⃗

=

⃗ ⃗

; ⃗

=

independent of ⃗

Therefore, Electric field within the cavity is constant (uniform) and depend on distance from the centre and always parallel to line joining centre of cavity and sphere, towards centre of sphere. ⃗ = 0 for concentric cavity. For Infinite linear charge: To determine field near an infinite linear charge of charge density at distance ‘r’ a coaxially cylindrical Gaussian surface is assumed that passes through the point of consideration. + + +  + + + + + + + +

+ +  0 + top + d s + + d s + + + + ds +  Bottom  0 + + + +

For infinite charged linear conductor, Electric field is away from axis and so is the direction of area vector of curved surface. The point to be noted here is that the top or bottom area vectors are axial hence ⊥ to electric field. Therefore flux generated by top and bottom will be zero. hence; ∅ = ∅

= ∮ ⃗. ⃗ =

⃗ = constant and ⇒ .2

=

=

⇒ ∅ = .∮

; sin =

⃗ and

;

; =

Note: For semi-infinite wire: Ey 

 4 0 r E

 2 4 r

Ex 

 4r

45 0

 

+ +

+ +

+

+

+

 +

+ +

+ +

+ +

+ +

+

For Cylindrical shell: To determine Electric field near cylindrical shell of radius R and uniform surface charge density , at a point r from axis. A coaxially cylindrical Gaussian surface passing through point of interest is considered. ⃗. ⃗ = A) inside the shell r < R; ⇒ = 0 { = 0} R + + + + + + + +

+

+

+

+

+ + +

+ + +

+ + + + + + + +

B)

; ⇒ ∮ ⃗. ⃗ =

Out side the shell ≥ ; Qin = . A = . 2 since ⃗ | | ⃗ and ⃗ = constant; . ∮ ; ⇒

=

⇒; . 2

=

= R + + + + + + + + +

+ + + + + + + + +

For uniform cylindrical charge distribution: To determine Electric filed near uniform cylindrical charge distribution, a coaxially cylindrical Gaussian surface is considered that passes through point of interest. Like above cases, Electric flux is generated by curved surfaces only. If charge density is and radius is R A) Inside r < R Qin =

; ∮ ⃗ . ⃗ =

since ⃗ | | ⃗ and ⃗ = constant; . 2 R

r l

=

; =

B)

Out side ≥ ; ; ∮ ⃗ . ⃗ =

Qin =

= constant; ⇒ . 2

since ⃗ | | ⃗ and ⃗

=

; = R

Note: For cylindrical cavity at distance d from axis and radius ⃗ ⃗ = ⃗ − ⃗

cavity

= ⃗=

− ⃗

⃗ ⃗

again uniform like spherical cavity

SESSION - 9 AIM  Electric field due to Infinite sheet of charge  Electric field due to Infinite charged conducting surface  Distribution of charge on Isolated conducting Infinite sheet of charge (Earthing case will be discussed with capacitor) THEORY 1] To determine Electric field due to infinite sheet of Charge: Sheet of charge refers to plane sheet, when point of interest is very close to sheet and away from the vertices and edges. Let’s consider a sheet of uniform surface charge density . The Gaussian surface will be equidistant from the sheet containing the point of interest. + E

+ +

+ +

+ +

+ + + + + + + + + + + + + + +

S

E

S S

From figure you can see, the direction of electric field is parallel to area vector Δ ⃗ Δ∅ = Δ∅ + Δ∅ = ⃗ . Δ ⃗ + ⃗ . Δ ⃗ = 2 Δ and

charge

2 Δ =

enclosed

; =

is

Δ = σ Δ S

hence;

Δ∅ =

;

2]

To determine the Electric field due to charged conducting surface: Here we mean a surface which is not flat. For such conducting surface all the charge is drifted to outer surface and Electric field is outwards only (+ ve charge outwards, ‘– ve charge inward direction) Therefore flux is considered on outer surface because for inner surface ⃗ = 0 So ∅ =

; .Δ =

; E

For Spherical surface

3]

 

 0

Q

4 R 2 Q E  4  0 R 2



For Cylindrical Surface Q

2 rL Q E 2 0 rL

Charge distribution on Infinite conducting sheets of charge Q1

Q2

Q3

Let initially charges given to sheets be Q1, Q2, Q3, Since charge comes on outer surfaces of conductor, no charge will remain within this conductor. If we choose a Gaussian surface through the plates, Electric field / Flux for these surfaces will be zero and total charge within the surface must be zero. Total charge will come on outer surfaces. We can directly consider induction also. Both the ideas will fetch same result but induction is faster.

Q1 Q1 2

Q3

Q2 Q1 2

Q2 2

Q2 2

Induction 

Q2 2

Q3 2

Q Q2  1 2 Induction 2

Q3 2

Q3 2

Q1 2

Q3 2

Q3 2 Induction 

Q3 2

Q2 2

Q2 2

Q1 2

Q1 2

Net Charge Q1  Q2  Q3 2

Q1  Q2  Q3 2

 Q1  Q2  Q3 2

Q1  Q2  Q3 2

 Q1  Q2  Q3 2

Q1  Q2  Q3 2

or similarly

1] 2] 3]

You can verify the result by checking following concepts. Total charge on inner surfaces, facing each other will be zero. Net charge on each plate is still same. Net charge on outer surfaces of any pair is the charge initially given to plates.

SESSION - 10 AND 11 AIM  To explain the concept of potential energy and work done.  To discuss Potential Energy of a system of point charges.  To develop concept of Potential and its relation with electric field.  Equipotential surfaces. THEORY Potential energy: Potential energy is the property guided by conservative forces. Since the Electrostatic Electric field (not induced electric field of Electromagnetic Induction) is conservative, hence the work done against Electrostatic force will be conserved as potential energy of the system. As per definition of conservative force: =

;

= −

.

Now consider a system of two charges q1 and q2. Let the charges be at infinite separation in the begining. (r = ∞; F = 0 ⇒ = 0) The charges are finally brought to a finite separation ‘r’ . The work done against conservative force i.e. negative of Wconservative will be stored as potential Energy of system. Hence, The force at some distance r; = and “Potential energy is defined as the work done required to bring two charges from infinity to a finite separation.”

⇒ ∫ −

=∫ −

; − 0 =

=

;

;

=

=0

Further more, work done required to change the separation from r1 to r2 Δ

= Δ ; =

; =

This is the property of conservative force. “The work done under the influence of conservative force is always independent of the path taken”. It is expressed in terms of displacement along the conservative forces only i.e. initial and final positions only”. Note: Unlike force and electric field, Potential energy and potential are scalar. So use the values of charges along with sign which is important to describe nature of the two. Potential Energy of system of charges: The equation discussed above is true for only two charges. For more than two charges we have to consider P.E. of all the possible pairs without repetition. The number of pair can be calculated as NC2 = C (N, 2) = ( ) ! = )! !( If number of charges is more then we should first calculate number of possible pairs from the formula above and then look for all pairs systematically. And then using = for all pairs we have to add them. Don’t forget that in this formula we have to account for positive negative sign of charges.

CLASS EXERCISES 1. Eight charges, q each are kept at vertices of a cube of edge length a. Find net potential energy. Sol: If you see the number of pairs, the number of pairs is × = = = 28 pairs seem complicated. Take another view. 12 edges of length = a; 12 Face diagonals of length = √2 (Two on each face) 4 Body diagonals of length = √3 Total Pairs = 12 + 12 + 4 = 28 Now you can write easily. =

12 × + 12 ×

+4

;=

3+

+

Concept of potential: Electric Potential: Electric potential at a point is something that describes P.E. of unit charge. By definition “Potential at a point in Electric field is the amount of work done to bring a unit charge from infinity to that particular point in Electric field.” i.e., = = ; Relation of Electric field and potential. Since we know

= − ∫ ⃗. ⃗

----(1)

Hence dividing first equation by q;

= ; ⃗ =

and

= −∫

. ⃗ ; =

− ∫ ⃗ . ⃗ ; This rule is very important because we have already seen Electric field for various Cases. Now use this formula and final out potential for all cases.

Spl. Cases: 1] If ⃗ =

̂+

. or ∫ Δ = −∫ . 2]

If

̂+ −∫

= ( , , ) then ⃗ =

⇒∫ .

−∫ ̂ −

= −∫

.

−∫

.

. ̂−

is partial derivative with respect to x, and to calculate this y and z are to be assumed as constant. And the same rule holds for other partial derivative with respect to y and z. Equipotential surfaces: The surface containing all the points in space at same potential is called equipotential surface. Note: * Conducting spheres and cavities always behave as equipotential surfaces. * Any charge induced due to presence of any external charge does not changes its potential because net induced charges is always zero. * Net force on the charge by shell or cavity is equal and opposite to force by external charge in absence of shell or cavity in opposite direction. Hence net force on the charge in conducting cavity or shell will be zero. This effect is known as “Electrostatic shielding”. Rules for Equipotential surfaces: 1] Electric field is always normal to these equipotential surfaces. 2] Like lines of electric field, two equipotential surfaces are never intersecting. (same reason)

Examples of Equipotential surface: 1] Point charge / shell / Sphere; ⇒ Concentric spherical surfaces.

2] 3] 4]

Linear charge / cylindrical; ⇒ Coaxial cylindrical surfaces. Sheet of charge / conducting planes ⇒ Equidistant planes. Two like charges

5]

Two unlike charges

6]

Three like charges +

+ +

SESSION - 12 AIM  To discuss potential due to different types of charge distribution a) Point charge b) Shell / Conducting sphere c) Uniform spherical charge distribution d) Ring e) Linear charge f) Cylindrical shell or conducting cylinder g) Charged disc h) Infinite sheet of charge  Graphical Representation of Electric field and Potential for charge distributions. THEORY Potential can be found directly as well as using electric field. In practice we determine Potential directly for the charges following Colombian Approach and with the help of Electric field for charges following Gaussian Approach. a) Point Charge: =−

=−

4

b) Shell / Conducting sphere: Inside r < R

− = 4

1

=

4

Q

R r

E = 0 ⇒ dV = 0 Potential within the shell is same as the value at surface. Q R r

Outside( r ≥ R)

= −∫

=

{same as point charge)

Potential inside = Potential on surface = c)

.

Uniform spherical charge distribution: Potential outside the sphere is same as the point charge only for ≥ ;

=

Inside the sphere( r < R) the field within the sphere is of different type (E

r) and outside is different

. Therefore we

integrate in two parts i.e. outside and inside separately. Inside Ein 

Qr 4 0 R3

r

R

outside = − ∫ =

.

=

=

=

[2

−∫

∫ −

−0−

. −

+ +

∫ ;

]=

;

d) I)

Ring: At centre - Every charge element is at a distance equal to the radius (R) of Ring. dQ

R

Charged Conducting Sphere or Shell: So

=

=

II) At Axis: Every charge element is at a fixed distance (x2 + R2)1/2 from the point. So again dQ x 2  R2

x

e)

Linear Charge: Here we will again follow same approach as we did for electric field. Q r

dx

(-42, 0)

x = d tan

= =

[log(

r = d sec

dQ = dx

=

+

=

= d sec2 d

;

sec

4

(-42, 0)

=

dx = d sec2 d

hence dV=

d



4

|log[

+

+ ]| −

) − log[sec(− ) + tan(− )]]

=

+ −

log

4

=

log

/

/

/

/

Note: for infinite charge, i.e. L >>>d we get

. So we cannot

determine potential at a point for infinite linear charge. But we can calculate Δ = (

Δ = −∫ + f)

ln

; = −

) = −

.

=−

=

Uniformly charged Cylindrical Volume: (I) Outside Δ =

g)

>

ln

like infinite linear charge

On the axis of charged disc: =−

= −

1−

2

+

Let x2 + R2 = t2; 2x dx = 2 + dt x; = ∞ ⇒ = ∞ = = i)

⇒ = √

+ +

=

−∫

;

Infinite sheet of charge: Again we cannot determine potential of point but near the surface Δ can be determined r2 r1

V  E r   E ( r2  r1 ) Since E is uniform near the surface

Infinite Sheet  V   (r2  r1) 2 0

2]

Conducting surface  V  ( r2  r1 ) 0

Graphs for Electric field and Potential: Electric field

Potential

E E

Point Charge (Q+)

Charged Conduct ing Sphere or Shell Uniform charge spherica l Distribu tion

V

1 r

+ve direction

2

-r

V

1 r

r Q

r Q

-ve direction

+

-E

Q

Q+ R

V the direction E

V

1

Q 40R

1 r

V

r2

Eo - ve direction Q R

r

E W

the direction E

1 r2

V r

- ve direction

2 2

V

1 r

V

Q

(3 R r )

8 0R3

V

1 r

E r

R/ 2

Ring

R/ 2

direction

V

+ve direction

r

V

Q 40R

V

Q 40 R2x2

SESSION - 13 AIM  Dipole and dipole moment.  Electric field and potential due to dipole. a) On Axial line or End on position; b) On Equitorial line or Broad on Position c) Arbitrary Position for a short dipole  Force between two dipoles (short)  Behavior of a dipole in Electric field A) Uniform Electric field. i) Force and Torque; ii) Potential Energy and work done B) Non-uniform Electric field. THEORY: Electric Dipole: Real dipoles are covalent molecules, that we study in chemistry (as discussed earlier). Here dipole means “two equal and opposite charges separated by a small distance”. 2l

-q

p

+q

Dipole Moment: Dipole moment is a vector directing from “– ve” charge to “+ve” charge. Its magnitude is the product of amount of either charge and separation, ‘–q’ to ‘+q’. ⃗ = 2 ⃗

Electric field and potential: a) Axial line or End on Position: Distance is considered from center of Dipole. r+l E

P -q

E

r-l

+q 2l r

You can see from figure that E+ due to positive charge is more than E– due to negative charge because ‘+ve’ charge is a little bit closer as compared to ‘– ve’ charge. Therefore ⃗ = ⃗ + ⃗ – = =

(

= =

− – due to apposite direction − = [4 ] ) ( ) ( ) . (

)

(

)

Note: ⃗ is parallel to dipole moment. =

r > > > l short dipole ⇒

Since potential is scalar, comes with sign of charge. So, = = ⇒

+ + − ; = (

)

;

(

=

+ )

(

= )

. (

)

for r > > > l short dipole.

b)

Equatorial line or Broad on position: In this case point lies on perpendicular bisector. Hence point is always equidistant from the charges, Electric field will be equal in magnitude but not on same line.  

E

E

E

q

l

l

⃗ = ⃗ + ⃗ But ⃗ + ⃗ E = 2E+ cos {law of parallelogram for two equal vectors) = 2 = =

c)

.

(

) (

(

) /

) /

Note: direction is antiparallel.

for r > > > l short dipole ⇒

Since point is equidistant from equal and opposite charges, therefore V = 0 at any point in equatorial plane. Arbitrary Position for short dipole: Now consider an arbitrary position i.e. at distance r making an angle with the dipole moment. E E1

p// p

+

900

p

+

+

E //

In this case we resolve dipole moment i.e. parallel to separation ⃗// and ⊥ to separation ⃗ = Electric field of ⃗// is axial while for ⃗ is equatorial hence 2 2 ⃗ = + = + = // ⊥ +4 √ hence

=

at an angle

√1 + 3 =

//

=

=

from

figure. if

=0⇒

Axial case if

= 900 ⇒ E =

case Special case for ⃗ ⊥ ⃗ E E //

E1

 p

q

From Figure: +

= 90 ;

tan(90 − ) = =

= 90 − ; Since tan 2

√2 =

; √

;

= 2; ;=

Therefore cos = ; So, = √

=

; = sin_1 √

when ⃗ ⊥ ⃗

Equatorial

3]

Force between two small dipoles: For small dipoles, length of two dipoles is very small. So we can consider it as dr. Case - 1 Dipoles on same axis. q 2 E1  dE1 q2 E1 q1 p1 q2

 q2 p2

 q2

you can see here field at –q2 is more than + q2 by dE1 magnitude. So net force for parallel dipole moments is of attraction (Repulsion of anti-parallel dipole moments) Electric field of 1st dipole at 2nd dipole E1=

|dE1| =

dr

Force on 2nd dipole is q2. dE1 F = q2 dE1; =

.

but q2 dr = p2

Radial Force. If ⃗ | | ⃗ ⇒ attraction;

Hence =

antiparallel ⇒ Repulsion. Case - 2 Dipole perpendicular to Each other. -q 1

p3 E

q(E + dE)

p1 -q2  q1

+q2

qE

Field of First dipole on second. E=

;|

F = q2 dE;=

|=

.

; net force on dipole-2 =

; tangential.

4] Dipole in Electric field: A) Uniform Electric field: Dipole is making an angle electric field as shown in figure. q

with the

qE 2l sin 

-qE

2l

p  2l cos 

Net Force: The force on ‘+ve’ charge is qE along Electric Field. The force on ‘-ve’ charge is qE opposite to electric Field. Hence Net force = 0 FN = 0 (dipole in uniform Electric Field) Net Torque: As you can see in figure the forces on dipole are making a couple separated by a distance of 2l sin . Hence net torque = F.2l sin = E q 2l sin = = pE sin OR ⃗ = ⃗ × ⃗ Potential Energy: Writing potential energy requires a reference. For a dipole U= 0 at = 900 i.e. Potential Energy is written with the reference when ⃗ ⊥ ⃗ . Since here torque is conservative (like force) or restoring, therefore dU =− ⃗. ; ∫ = −∫ ⃗. ⃗ From figure you can see, angle is anticlockwise i.e., ‘+ve’ but Torque is clockwise i.e. ‘-ve’, So ⃗. ⃗ = − ] ) ; = [− ⇒ = − ∫ (− 90 = – pE cos + 0; ⇒ =−

= − ⃗. ⃗

=0⇒

= 90;

⃗ opposite to ⃗ ; Unstable Equilibrium; Umax = pE⇒ = 1800 Umin = – pE⇒ = 0; ⃗ parallel to ⃗ ; stable Equilibrium. B)

W = ΔU = U2 – U2 = pE (cos − ) Dipole in Non Uniform Electric Field: You can see Torque is again non zero ⃗ ≠ 0 but here force is also non zero ⃗

≠ 0 (unlike uniform

Electric Field) From figure E1 q

qE1 cos1

1 qE

qE1 sin1 qE 2 sin 2

p

qE 2

qE 2 cos 2

2

E2

F = q E1 cos _ q E2 cos = (qE1 sin + q E2 sin ) l {about CM} = (qE sin ) 2l { about ‘-ve’ charge) = (q E sin ) 2l { about +ve charge)

CLASS EXERCISE: 1] If the distance of the point from a source generating electric field is doubled what is the effect on the magnitude of electric field for the following 2 casesA] If the source is a point charge B] If the source is a small electric dipole 2] What is the ratio of magnitude of electric field created by a very small electric dipole at equal distances for broad side and end-on position? 1] An electric dipole of dipole moment p is placed parallel to uniform electric field E. Find the work done by the external agent in rotating the electric dipole from this position by an angle of a] 600 b] 900 c] 1800 2] An electric dipole of dipole moment p is placed in a uniform electric field E such that p is parallel to E. Find the following. a] Net force experienced by the electric dipole. b] Net torque experienced by the electric dipole. c] Is it a case of equilibrium. If yes, then define the nature of equilibrium. 3] Repeat the above problem such that the dipole moment p is antiparallel to direction of uniform electric field E.

SESSION - 14 AIM: 

   

Use of energy density to calculate energy of formation, within the shell & sphere of uniform charge distribution and outside these systems. Electrostatic pressure on shells. Force required to hold two hemispherical shells having same charge density together. Charge distribution / Induction on spherical concentric shells. Charge distribution on irregularly shaped conductor. and “Corona Discharge”

THEORY: 1] Energy density is a term, new for us at this stage and will be discussed in capacitors but its very important application is here. So let’s start. Energy density in electric field is u =

i.e. Electrostatic

Energy stored (PE) per volume. We will use this definition to determine Energy of a uniformly charged solid sphere. Consider a uniformly charged sphere of charge Q and radius R. Energy stored outside the sphere will be a function of Electric field outside the sphere. =∫ ;

.

; ℎ

=

=4

; =

=∫ −

.

.4

;

=

; ;

=

=0

Energy stored within the sphere =∫

here

=

and

=4

;

=

.4

∫ =

; =

=

So Energy of Assembling the charge i.e. total energy of uniformly charged sphere U = Uin + Uout =

+

=

(

)

; =

Note: 1) For spherical shell, Ein = 0 ⇒ Uin = 0 hence

=

(Outside the sphere only)

2) Total energy can be calculated by other method also. Let us see how ! Charge density of sphere = = .

; =

and

; So charge in the radius

=3=

.

=

Potential due to this charge on a sphere of radius r only is =

=

to set an additional charged shell around

the sphere of radius r; dW = V dq ⟹

=∫

=∫

.

=

=

2]

Same Result, but here we cannot distinguish Energy within the sphere and outside the sphere. Consider any shell of uniform surface charge density . Its behavior will be like conducting surface only. Now think of a hole of area ds. The field near the surface will be = at any point near the surface + + + + + + + + + + + + + + + + + + + +

+ + + + + + + + + +

E0

ds

E' E0

E'

The removed portion has a field at both the sides. The field near hole is E0 just within and just outside the shell due to remaining portion of shell. Since in the shell near the hole Enet = 0 ⇒ E0 = { both opposite within the shell. Out side the shell near the hole Enet = ⇒ + 2 = (

=

)⇒

=

Charge on the removed part. dq = Force on removed part. dF = E0 dq;=

So pressure

experienced by removed portion when it was the part of shell. =

=

= .

=

Note: Electrostatic Pressure in ⃗ External is = ⊥ for surface of charge density.

So Electrostatic pressure at point on shell (surface) =

3]

=

Luckily the relation is similar to Energy density in dimensions but physical content of quantity is totally different. Force required to hold two hemisphere of similar charge densities + +

+

+

+ +

R

+ + +

×

=

.

+ + + +

+ + + + +

Electrostatic Pressure = FRepulsion =

+

Force of Repulsion =

So Force required to hold the hemisphere F = FRepulsion = The two hemispheres can also be held by placing a charge –Q at centre. Field due to charge =

;

Since Field is parallel to Area Vector. So pressure = =

=

; = 2

; hence FAthaction = FRepulsion ; =

/2 of ‘– ve’charge

4]

Charge distribution / Induction on spherical conducting shells. A) One shell under influence of External charge Q:q + + + + +

Q r

Here shell has charge q and a charge Q is placed at a distance r from its centre. Potential due to charge q on the shell itself Vindividual = Potential due to External charge Q VExternal =

2] 3]

; Vnet = Vindividual + VExternal; =

+

The point to be noted here is that lines of electric field must always normal to conducting surface while incident or emerging out of the surface. The point of incidence signifies induction of negative charge and point of emergence signifies induction of positive charge. If shell is earthed, its potential will become zero. The final charge on the shell is the charge induced to keep its potential zero. qin

R

Q

r

So Vnet = 0;

+

= 0 ; qin = −

The charge induced is

=

and it is again distributed

uniformly if shell is again made isolated. qin

R

B]

Two concentric Conducting shells q2 q1

b a

As you see in figure, charge q1, is given to inner shell.  q1  q1 + + + q1 + q2 + + + + + b

You can see that total charge of inner shell is induced on inner surface of outer shell as – and successively on outer surface of outer shell as+ . Note worthy point here is that net charge on any shell is still same. So potential on inner shell Viner = due to charge q1 on it and Vouter =

due to charge q2 on

external shell. Net Potential V1= Vinner + Vouter = Similarly for outer shell Vouter = shell Vinner=

due to charge q2 on outer

due to interior shell of charge q1.

Net potential V2 = Vinner + Vouter =

+

Now if inner shell is earthed. Final charge on inner shell be q11. Potential or inner shell becomes zero due to earthing. q11

b

So

=

+

= 0; ⇒

 q11  q11  q2

=−

;

=−

Find charge distribution + + + +

+

+ ba + b

+ + 

+

+ + + +

q2

a q2 b +

a q 2  q2 b

+ + ba +   b q2   +

External surface of outer shell. Similarly charge on bigger shell be 4 ⇒ = Now

;

= =

= ; where

&

are respective charge

densities. So charge density (surface) is more for smaller radius and vice verse. Therefore Field near smaller sphere = and Field near bigger sphere

=

; So

>

because

>

Hence, Possibility is higher for corona discharge near smaller sphere. This is the reason why “Lightning Conductors” are installed on tall buildings. These Lightning conductors are in the form of spiker, to have higher electric field due to smaller radius at tips. Therefore they can absorb all the electrical energy of lightning and conduct it safely to ground with out damaging the building.

Suppose now external surface of outer shell is earthed. The charge on outer surface is zero. q12

b q1

The charge on inner surface of outer shell inner shell = Potential of outer shell = 0 = 0; =

= 0 ; ⇒

=−  q1 +

+ b

+ +

5]

. The charge on

+q 1 + +

+

Final charge distribution you can repeat the process for any no of shells. For irregular shaped conductor the potential at the surface remains uniform but charge density becomes different. This difference in charge density produces variation in Electric field near the surface. When this electric field becomes greater than or equal to dielectric Strength i.e. the maximum possible electric field that a medium can sustain, ionization of medium (dielectric) starts resulting in dissipation of charge from the surface (Sparking). This process is know as corona discharge.