Crackiitjee.in.Phy.ch16
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ELECTRIC FIELD Example: A ring-shaped conductor with radius a carries a total positive charge Q uniformly distributed on it. Find the electric field at a point P that lies on the axis of the ring at a distance x from its centre. Solution:
e.
in
The figure shows the electric field dE at the point P due to an infinitesimal segment at the top of the ring. The x-and y-components of dE is dE cos Ө and dE sin Ө. If we take another segment of the bottom, we see that its contribution to the field at P has the same x-component but opposite y-component. Hence total y-component becomes zero.
=
=
dE cos
cr ac k
E=
iit je
Hence the field at P will be along x-axis. Therefore, the net field at P is
1
4
.
0
dQ x . 2 x a x2 a2 2
x
3 2 2
2
40 x a
dQ
Qx
=
3 2 2
2
40 x a or, E
Qx
2
3 2 2
40 x a (1)
i
At the centre of the ring, x = 0 , E = 0
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(2)
If x >> a, E
1 Q i 40 x2
This field is same as that of a point charge at the centre. E will be maximum at the points where or, x =
a , 2
and Emax =
1 2Q . 40 3 3a2
dE =0 dx
in
(3)
iit je
e.
The graph of the variation of E with x is as shown in figure.
The negative value of E indicates that the direction of the field is opposite to the positive value. If a >> x, E=
=
cr ac k
(4)
Q x 40a3
x, 20a2
where is the linear charge density. The force on a negative point charge –q is F=
q x 20a2
= – kx, where k =
q 20a2
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On releasing the point charge (–q) of mass m, it will execute SHM (if x 0 if Vi > Vf. Hence a positive charge moves from high potential to low potential if it is free to move. For a negative charge, W > 0 if V i < Vf. Hence a negative charge moves from low potential to high potential if it is free to move.
(viii)
The work done by the external force is W ext = q(Vf – Vi), (note the order of subscripts), if the charge q moves from i to f in a electric field very slowly.
cr ac k
iit je
e.
in
(i)
Potential due to a Point Charge It is the work done by an external agent in moving unit positive charge slowly from infinity to the point where potential is to be calculated. Figure shows an isolated positive point charge of magnitude q. We want to find potential V due to q at a point P, at a radial distance r from that charge.
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This potential will be determined relative to zero potential at infinity. Let us suppose that a test charge (positive) q0 moves from infinity to point P. Since the work done by the electric field is independent of the path followed by the test charge, we take the simplest path along the radial line from infinity to q passing through P. Consider an arbitrary position P' at a distance r' from q. The electrostatic force on q at P' is
F
1 qq0 , directed radially away. For a further differential 40 r '2
in
displacement ds, the work done by the force F is
e.
dW = – F.ds
Since the direction of ds is opposite to the direction of increasing r'.
iit je
Hence ds dr ' So, dW = Fdrc
1 qq0 dr ' 40 r '2
cr ac k
=
1 or, W = qq0 40 =W
r
dr ' r '2
r
1 1 qq0 = 40 r ' =
1 1 1 qq0 r 40
=
qq0 40r
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V
W q q0 40r
…[potential due to a point charge] If the point charge is negative the electrostatic force points towards q, and hence W will be positive, giving a negative potential V. Thus the sign of V is same as the sign of q, when we assign zero potential at infinity.
in
Potential due to an Electric Dipole
1 q 40 r1
cr ac k
V=
iit je
e.
Figure shows a dipole of dipole moment p = qd. We have to find the electric potential at the point P, distant r from the centre O of the dipole. The line OP makes an angle Ө with the dipole axis. The potential at P due to the positive charge (+ q) is
and that due to the negative charge (– q) is V2 =
1 q 40 r2
The net potential at P is V = V1 + V2
q q r r 1 2
=
1 40
=
q r2 r1 …(i) 40 r1r2
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For a small dipole, we have usually, r >> d, where "d" is the separation between the two charges. Under these conditions, We may write, r2 – r1 2 - d cos Ө and r1 r2 - r using these quantities in equation (i), we get V=
q dcos . 40 r2
Example: Infinite number of same charge q x = 1, 2, 4, 8 ...... What is the potential at x = 0 ?
are
placed
at
2q 40
=
q 20
Example:
cr a
=
ee
q 1 1 40 1 2
ck i
=
1 q q q q .... 40 1 2 4 8
itj
V=
.in
Solution:
If the alternative charges are unlike, then what will be the potential? Solution: Then, V=
1 q q q q ......... 40 1 2 4 8
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1 q q q q ..... ..... 40 1 2 2 8 1 1 1 1 1 2 1 1 4 4
=
1 40
=
1 2q 40 3
in
=
e.
Potential difference:
iit je
The work done in taking a charge from one point to the other in an electric field is called the potential difference between two points. Thus, if w be work done in moving a charge q0 from B to A then the potential difference is given by–
(i) (ii) (iii) (iv)
W q0
cr ac k
VA – VB =
Unit of potential difference is volt. This is a scalar quantity.
Potential difference does not depend upon Co-ordinate system. Potential difference does not depend upon the path followed. This is, because electric field is a conservative force field and work done is conservative force field does not depend upon path followed.
Example: In the following fig. Along which path the work done will be maximum in carrying a charge from A to B in the presence of any another charge.
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Solution: Same for all the path [Because the work done doesn't depend upon the path]
Example: A charge 20 µC is situated at the origin of X-Y plane. What will be potential difference between points (5a, 0) and (–3a, 4a)
Distance between (0, 0) & (5a, a),
25a2 0
e.
r1 =
kq 5a
cr ac k
Distance between
iit je
= 5a
V1
in
Solution:
(0, 0) & (–3a, 4a) r2 = = 5a V2 =
9a2 16a2
kq 5a
V1 – V2 = 0
Relationship between electric potential and intensity of electric field A
(i)
VA = E.dr,
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VA = electric potential at point A. (ii)
Potential difference between two points in an electric field is given by negative value of line integral of electric field i.e., B
E.dr
VB – VA =
A
(iii)
E V = – grad
= (gradient)
Ey =
V , y
iit je
V , x
cr ac k
Ex =
in
e.
= i j k x y z
Ez =
V z
(iv)
If v is a function of r only, then E =
dV dr
(v)
For a uniform electric field, E =
V and it's direction is along the r
decrease in the value of V.
Example: Electric potential for a point (x, y, z) is given by V = 4x 2 volt. Electric field at point (1, 0, 2) is –
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Solution: E=
dV dx
= – 8x E at (1, 0, 2) = – 8 V/m Magnitude of E
Example:
Solution:
dV dx
cr ac k
E=
e.
and x = 20 m will be
100 potential difference between x = 10 x2
iit je
Electric field is given by E =
in
= 8V/m direction along –x axis.
dV Edx B
B
A
A
dV E.dx 20
100 x2 10
VB VA = – 5 volts
Potential difference = 5 volt.
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Example: The
potential
at
a
point
(x,
0,
0)
is
given
as
V
=
1000 1500 500 x x2 x3 . What will be electric field intensity at x = 1m ? Solution: E=–V
V V V j k x y z
in
= i
V x
cr ac k
=
V V V j k x y z
iit je
=i
e.
or iEx + jEy + kEz
V V y 0 z
Comparing both sides Ex = =
V x
1000 1500 5000 x x x2 x3 1000 2 1500 3 5000 x2 x3 x4
=
For x = 1, (Ex) = 5500 V/m
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Example: What will be the electric field intensity at r = 3. Solution: For 2 < r < 4, V = 5 volts
dV 0 dr
in
E
e.
In the above problem, what will value of E at r = 6 ? at r2 = 7m
iit je
V2 = 2 volt at r1 = 5m
cr ac k
V1 = 4 volt
V V
1 E= 2 r2 r1
2 4 7 5
=
= 1 volt/metre.
Example: An oil drop 'B' has charge 1.6 x 10–19 C and mass 1.6 × 10–14 kg. If the drop is in equilibrium position, then what will be the potential diff. between the plates. [The distance between the plates is 10mm]
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Solution: For equilibrium, electric force = weight of drop qE = mg
V mg d
V
mgd q
1.6 1014 9.8 10 103 = 1.6 1019
e.
V = 104 volt
in
q.
(i) (ii) (iii) (iv)
cr ac k
Equipotential Surface –
iit je
[When a charged particle is in equilibrium in electric field, the following formula is often used qE = mg]
These are the imaginary surface (drawn in an electric field) where the potential at any point on the surface has the same value. No two equipotential surfaces ever intersects.
Equipotential surfaces are perpendicular to the electric field lines. Work done in moving a charge from a one point to the other on an equipotential surface is zero irrespective of the path followed and hence there is no change in kinetic energy of the charge.
(v)
Component of electric field parallel to equipotential surface is zero.
(iv)
Nearer the equipotential surfaces, stronger the electric field intensity.
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Example: Some equipotential surfaces are shown in fig. What is the correct order of electric field intensity?
Solution: EB > EC > EA, because potential gradient at B is maximum.
POTENTIAL ENERGY OF CHARGED PARTICLE IN ELECTRIC FIELD Work done in bringing a charge from infinity to a point against the electric field is equal to the potential energy of that charge.
(ii)
Potential energy of a charge of a point is equal to the product of magnitude of charge and electric potential at that point i.e. P.E. = qV.
(iii)
Work done in moving a charge from one point to other in an electric field is equal to change in it's potential energy i.e. work done in moving Q from A to B = qVB – qVA
ck i
itj
ee
.in
(i)
(iv)
cr a
= UB – UA
Work done in moving a unit charge from one point to other is equal to potential difference between two points.
NOTE: Circumference of the circle in above example can be considered as equipotential surface and hence work done will be zero.
Potential Energy of System:
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(i)
The electric potential energy of a system of charges is the work that has been done in bringing those charges from infinity to near each other to form the system.
(ii)
If a system is given negative of it's potential energy, then all charges will move to infinity. This negative value of total energy is called the binding energy.
(iii)
Energy of a system of two charges PE =
(iv)
1 q1q2 40 d
in
Energy of a system of three charges
e.
1 q1q2 q2q3 q3q1 40 r12 r23 r31
iit je
PE = (v)
Energy of a system of n charges.
n qj j1 rij i j
cr ac k
1 1 n PE = . q 2 40 i1 i
Electric Field Example:
Two symmetrical rings of radius R each are placed coaxially at a distance R meter. These rings are given the charges Q 1 & Q2 respectively, uniformly. What will be the work done in moving a charge q from center of one ring to centre of the other. i Solution: Work done = q (V2 – V1) potential at the centre of first ring
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V1 =
=
1 Q1 1 40 R 40
Q2 R2 R2
Q 1 Q1 2 40R 2
potential at the centre of second ring V2 =
Q1 R2 R2
work done = q (V2 – V1 )
Q Q q Q1 2 Q2 1 40R 2 2
q 1 Q1 Q2 1 2 40R
cr ac k
W12
iit je
=
in
Q 1 Q2 1 40R 2
e.
=
1 Q2 1 40R R 40
NOTE:
Work done in moving the same charge from second to the first ring will be negative of the work done calculated above i.e.
W21
q 1 Q1 Q2 1 40R 2
ELECTRIC DIPOLE (i)
A system consisting of two equal and opposite charges separated by a small distance is termed an electric dipole.
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Example: Na+ Cl–, H+ Cl– etc. (ii)
An isolated atom is not a dipole because centre of positive charge coincides with centre of negative charge. But if atom is placed in an electric field, then the positive and negative centres are displaced relative to each other and atom become a dipole.
(iii)
DIPOLE MOMENT:
.in
The product of the magnitude of charges and distance between them is called the dipole moment. This is a vector quantity which is directed from negative to positive charge.
(b)
Unit:
itj
ee
(a)
Coulomb – metre (C-M) Dimension:
ck i
(c)
[M0 L1 T1 A1]
It is denoted by p that is p qd
cr a (d)
Electric field due to a dipole (i)
There are two components of electric field at any point (a)
Er in the direction of r
(b)
Er in the direction perpendicular to r Er =
1 2P cos . 40 r3
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E = (ii)
Resultant
Er2 E2
E= = (iii)
q P sin . 40 r3
P 1 3 cos2 3 40r
Angle between the resultant E and r, is given by
E
itj
If Ө = 0, i.e. point is on the axis – Eaxis =
1 P . 3 40 r
ck i
(iv)
ee
1 tan = tan 2 –1
.in
Ө= tan–1 Er
(v)
cr a
Ө = 0, i.e. along the axis.
If Ө = 90°, i.e. point is on the line bisecting the dipole perpendicularly Eequator =
1 P . 3 40 r
(vi)
So, Eaxis = 2Eequator (for same r)
(vii)
Eaxis =
1 2Pr . 40 r2 2
2
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Eequator =
1 2P . 40 r2 2
3 /2
where P = q . 2 V=
1 40
r2
1 P cos . 40 r2
=
1 P.r . 40 r2
=
1 P.r . 40 r3
e.
in
=
iit je
(viii)
q 2 cos .
where Ө is the angle between P and r.
(x) NOTE : (i) (ii)
If Ө = 90°, Vequator = 0
cr ac k
(ix)
Here we see that V = 0 but E 0 for points at equator.
This is not essential that at a point, where E = 0, V will also be zero there eg. inside a uniformly charged sphere, E = 0 but V ≠ 0
Also if V = 0, it is not essential for E to be zero eg. in equatorial position of dipole V = 0, but E ≠ 0
Electric Dipole In an Electric Field – Uniform Electric Field (i)
When an electric dipole is placed in an uniform electric field, A torque acts on it which subjects the dipole to rotatory motion. This Ʈ is given by Ʈ = PE sin Ө
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or P E (ii)
Potential energy of the dipole U = – PE cosӨ = P.E or, V
1 p cos . 40 r2
…*potential due to an electric dipole+
e.
Potential due to a System of Charges
in
where p = qd is the magnitude of the electric dipole moment.
cr ac k
The potential at P,
iit je
Suppose a system contains charges q1, q2, q3 ....... qn, at distances r1, r2, r3 ....... rn, from a point P.
Due to q1 is V1 =
1 q1 40 r1
Due to q2 is V2 =
1 q2 40 r2
Due to q3 is V3 =
1 q3 40 r3
and so on. By the superposition principle, the potential V at P due to all the charges is the algebraic sum of potentials due to the individual charges. Therefore, net potential at P is V = V1 + V2 + V3 + …… + Vn
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q 1 q1 q2 q3 ............. n 40 r1 r2 r3 rn
or, V
Example :
in
In the given figure, there are four point charges placed at the vertices of a square of side a = 1.4 m. If q1 = + 18 nC, q2 = – 24 nC, q3 = + 35 nC and q4 = + 16 nC, then find the electric potential at the centre P of the square. Assume the potential to be zero at infinity.
e.
Solution:
The distance of the point P from each charge is
1.4m 1.4
cr ac k
=
a 2
iit je
r=
= 1m
V = V1 + V2 + V3 + V4
q1 q2 q3 q4 r r r r4 2 3 1
=
1 40
=
1 q1 q2 q3 q4 . 40 r
9 1 18 24 35 16 10 C = . 40 1m
= (9 × 109) × (45 × 10–9) V = 405 V
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Example: Two point charges – 5 µC and + 3 µC are placed 64 cm apart. At what points on the line joining the two charges is the electric potential zero. Assume the potential at infinity to be zero. Solution:
e.
in
For the potential to be zero, at the point P, it will lie either between the two given point charges or on the extended position towards the charge of smaller magnitude.
iit je
Suppose the point P is at a distance x cm from the charge of larger magnitude. Then from figure (a)
cr ac k
1 5 106 3 106 . 0 40 x 102 64 x 102 or,
3 5 0 64 x x
or,
8x 320 0 x 64 x
or, x = 40 cm From the figure (b), we get
3 5 0 x 64 x
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or,
320 2x 0 x x 64
or, x = 160 cm
Potential due to a Continuous Charge Distribution
in
When a charge distribution is continuous, instead of consisting of separate point charges, we divide it into small elements, each carrying a charge dq. Considering this differential element of charge dq as a point charge, we determine the potential dV at the given point P due to dq, and then integrate over the continuous charge distribution.
1 dq 40 r
iit je
dV =
e.
Assuming the zero of potential to be at infinity,
Where r is the distance of P from dq.
cr ac k
The potential V at P due to the entire distribution of charge is given by
V dV
1 40
dq r
Case I : Line charge distribution
For the charge uniformly distributed along a line dq = dl, where is the linear charge density and dl is a line element. Case II : Surface charge distribution For the charge uniformly distributed on a surface with surface charge density , the charge on a surface element dA is dq = dA Case III: Volume charge distribution For the charge uniformly distributed in a volume with volume charge density , the charge in a volume element dV is dq = dV.
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Example : Electric charge Q is uniformly distributed around a thin ring of radius a. Find the potential at a point P on the axis of the ring at a distance x from the centre of the ring.
.in
Solution:
1 40
x2 a2
ck i
NOTE:
Q
itj
V=
ee
Since the entire charge is at distance r = x2 a2 from the point P, therefore, the potential at P due to the total charge Q is
1 Q 40 a
cr a
At the centre, V =
Various Cases:
A thin uniformly charged spherical shell (a)
Potential at an outside point: We already know that for all points outside the sphere, the field is calculated as if the sphere were removed and the total charge q is positioned as a point charge at the centre of the sphere.
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Hence, the potential at an outside point at a distance r from its centre is same as the potential due to a point charge q at the centre. i.e., V
1 q 40 r
…*for an outside point+ Remember: This is valid only when the charge distribution is uniform. Potential at an inside point:
in
(b)
We know that inside the spherical shell of uniform distribution of
cr ac k
iit je
e.
charge, the electric field E is zero everywhere. Hence, if a test charge q0 is moved inside the sphere no work is done on that charge against the electric force. Hence for any infinitesimal displacement anywhere inside the sphere, W = 0 or q0 V = 0 or, V = 0. Hence V (potential) remains constant inside the sphere and is equal to its value at the surface
i.e., V
1 q 40 R
…[for an inside point, r < R]
A Charged Conducting Sphere Since the charge on a conducting sphere is uniformly distributed on its surface, therefore, it acts just like a spherical shell of charge uniformly distributed in it. So V=
1 q 40 r
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...[for an outside point, r > R] and V =
1 q 40 R
...[for an inside point, r < R]
Important Points about Conducting Sphere: For a charged spherical shell Vinside = Vsurface = REsurface = constant
(2)
Voutside = rEoutside
in
(1)
(4)
iit je
Figure shows the variation of V with r for a positively charged spherical shell. V is maximum inside and on the surface of the sphere.
cr ac k
(3)
e.
It increases or decreases with increasing r depending on whether q < 0 or q > 0.
Figure shows the variation of V with r for a negatively charged spherical shell. V is minimum inside and on the surface of the sphere. It is maximum (= 0) at infinity.
(5)
The maximum potential (Vm) upto which a spherical shell can be raised is Vm = REm where Em is the electric field magnitude at which air becomes conductive (known as dielectric strength of air). Its value is about 3 × 106 N/C.
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Thus, a conducting sphere of 1 m radius can be charged upto a maximum potential of 3 × 106 volt in air.
Potential Due to Uniformly Distributed Charge in a Spherical volume Case I: At a point P outside the sphere:
1 q 40 r
itj
VP =
ee
.in
As we know for all external points, the total charge q is supposed to be concentrated at the centre of the sphere. So the potential at an outside point P, distant r from the centre O is
ck i
…potential at an outside point i.e., for r > R Case II:
cr a
At a point P inside the sphere:
Let us divide the entire sphere into two parts, one is a sphere of radius OP = r and the other is the spherical shell of internal radius r and external radius R. As in case I, the potential at P due to the charge of sphere of radius r is V1 =
1 q' 40 r
4 3 1 3 r = 40 r
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r 2 = 3 0 q
where
4 R 3 3
For finding the potential due to the other part, let us consider a thin spherical element of radius x and thickness dx. Its volume is 4πx2dx, and hence it contains the charge dq = (4πx2dx)p = 4πpx2dx The potential at P due to this element is
1 4x2dx = 40 x 1 xdx 0
cr ac k
=
e.
in
1 dq 40 x
iit je
dV =
The potential at P due to the spherical shell of internal radius rand external radius R is V2 = =
dV
xdx 0 R
x2 = 0 2 r =
R 2 r2 20
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So, the net potential at P is V = V1 + V2
2 2 r2 R r = 30 20
6 0
4 R 3 60 3
or, V =
3R
2
r2
q 3R2 r2 3 80R
in
q
e.
=
iit je
=
3R 2 r2
(1)
(2)
cr ac k
….[potential at an internal point i.e., r < R]
Putting r = R, we get Vsurface =
Putting r = 0,
we get Vcentre =
= (3)
q 40R
3q 80R
3 V 2 surface
Figure shows the Variation of V with r. For 0 < r < R, the graph is parabolic.
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For r > R, the graph is hyperbolic.
EQUIPOTENTIAL SURFACES "An equipotential surface is a surface on which the electric potential V is same at every point". The electric field in a region can be represented by a family of equipotential surfaces, each of which corresponds to different potential.
e.
in
Figure (a) and (b) show different arrangement of charges and corresponding field lines, and cross-sections of equipotential surfaces for (a) the uniform electric field due to a large plane sheet of charge (b) the electric field associated with a positive point charge.
Example :
cr ac k
iit je
Figure (c) represents the equipotential surfaces for an electric dipole. Figure (d) represents the equipotential surfaces for two identical positive charges.
Some equipotential surfaces are shown in the figure below. What can you say about the magnitude and the direction of the electric field?
Solution: First we will find the direction of the E noting that the E lines are always perpendicular to the local equipotential surfaces and points in that direction in which the potential is decreasing. So it will be as shown below
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Note that we have drawn the E lines perpendicular to all equipotential surfaces and it is pointing in that direction in which the potential is decreasing. So, the total angle E is making with the positive x-axis is 120° as shown.
Then calculate,
Change in potential Distance travelled
ee
E=
.in
Now, how to calculate the magnitude of the electric field! For this we need to jump from one equipotential surface to another through shortest (perpendicular) distance.
cr a
ck i
itj
Let us suppose we are jumping from 20 V line to 10 V line from B to A. Note that the BA line is the shortest length between the two equipotential surfaces. The potential difference between the two lines is (20 V – 10 V) = 10 volt. Now, calculate the length AB = 10 sin 30° [cm] = 5 cm = 5 × 10–2 m Therefore, E=
10 volt 5 102 m
= 200 [volt/m] Hence, the complete answer is that the E has the value 200 V/m and makes 120° with positive x-axis.
Example :
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The electric potential existing in space is V(x, y, z) = B(xy + yz + zx). Find the expression for the electric field at point P(1, 1, 1) and its magnitude if B = 10 S.I. unit.
Solution: The formula E =
dV in extended form is written as follows: dr
V V V , and x y z
e.
where,
in
V V V E i j k y z x
V , then other variables y and z will be treated as constant. x
cr ac k
calculating
iit je
are partial derivatives of potential V with respect to x, y and z respectively. For partial derivative you should note that if say you are
Now, If we write
E iEx jEy k Ez , then Ex =
V , x
Ey
V y
and Ez
V . z
In the question
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V = B(xy + yz + zx) So, Ex =
V x
= – B(y + z) because variables y and z are treated constant. Now, put the values of B, y and z. Given B = 10 and (x, y, z) are (1, 1, 1) respectively.
= – B(y + z) = – 10 (1 + 1)
cr ac k
= – 20 N/C
e.
V x
iit je
Ex =
in
So,
Similarly, Ey =
V y
= – B(x + z)
[now treating x and z constant] therefore Ey = – 10 (1 + 1) = – 20 N/C and Ez =
V z
= – B (x + y)
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= – 20 N/C therefore,
E Ex i Ey j Ezk
or, E 20i 20j 20k N / C Now, if we want the magnitude of E , then it's equal to
E2x E2y E2z
20
2
20 20 2
2
iit je
e.
= 20 3
in
=
Example:
cr ac k
An electric field E i20 j30 netwon/coulomb exists in the space. If the potential at the origin is taken to be zero, find the potential at (2m, 2m).
Solution:
We have E=
dV dr
so, it can be written in vector form as dV = E.dr Note, you can write E as Ex i Ey j Ez k
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and dr dxi dy j dzk therefore,
E.dr Ex .dx Ey .dy Ez .dz In the given question the z component of E or the point is not given. So you can write
E.dr Ex.dx Ey.dy Now,
iit je
or, dV = – 20 dx – 30 dy
e.
= 20i 30 j . dx i dy j
in
dV = E.dr
cr ac k
Now we will have to integrate it within limits. Given V = 0 when x = 0 and y = 0 (lower limit) and we have to calculate V when x = 2 and y = 2 (upper limit) therefore, V
dV 20 0
y 2
x 2
dx 30
x 0
dy
y0
or, V 0 20 x0 30 y0 V
2
2
or, V – 0 = – 20 (2 – 0) – 30 (2 – 0) or, V = – 40 – 60 = – 100 volt.
POTENTIAL ENERGY OF A SYSTEM OF CHARGES
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"The electric potential energy of a system of fixed point charges is equal to the work that must be done by an external agent to assemble the system, bringing each charge in from an infinite distance".
ee
.in
Let us first consider a simple case of two point charges q1 and q2 separated by a distance r. To find the electric potential energy of this two-charge system, we consider the charges q1 and q2 initially at infinity. When one of the two charges (say q1) is brought from infinity to its place (say A), no work is done, because no electrostatic force acts on it. But when we next bring the other charge q2 from infinity to its location (B), we must do work because no, exerts an electrostatic force on q2 during its motion.
1 q1 40 r
ck i
V=
itj
Assuming the potential at infinity to be zero, the potential at B due to the charge q1 at A is
cr a
The work done by the external agent in bringing the charge q2 from infinity to B is W = q2 V =
1 q1q2 40 r
So, the potential energy is U=W =
1 q1q2 40 r
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This formula U =
1 q1q2 is true for any sign of q1. If q1 and q2 are of 40 r
same sign, q1 . q2 > 0, so the potential energy U is positive. If q1 and q2 are of opposite signs, q1q2 < 0, and hence U is negative.
Now, we consider a three-charges system shown in the figure. First, we suppose that three charges are at infinity. Then we bring them one by one to their locations.
e.
1 q1q2 …(ii) 40 r12
iit je
W1 =
in
As before, the work done by the external agent in bringing the two charges q1 and q2 (one by one is)
Now, the potential at C, due to q1 (at A) and q2 (at B) is
1 q1 q2 40 r13 r23
cr ac k
V=
So, when q3 is brought from infinity to C, the work done on it by the external agent is W2 = q3V =
1 q1q3 q2q3 40 r13 r23
The total work done is W = W1 + W2 =
1 q1q2 q1q3 q2q3 40 r12 r13 r23
Hence, the potential energy of the system is U=W
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1 q1q2 q1q3 q2q3 40 r12 r13 r23
=
or, U = U12 + U13 + U23 This can be generalized for any system containing n point charges q1, q2, q3..........qn. The potential energy of the system will be written as U = (U12 + U13 + ………. + U1n) + (U23 + ……… + U2n) + (U34 + ……… + U3n) + …………. + U(n – 1)n
U
=
qi .qj
1
4 i j
in
ij
i j
0
rij
e.
or, U =
cr ac k
iit je
Thus, the potential energy of a system of charges is the sum of potential energies of air possible pairs of charges that can be formed by taking charges from the system.
POTENTIAL ENERGY IN AN EXTERNAL FIELD (i)
Potential Energy of a single charge: The work done by an external agent in bringing a charge g from infinity to the present location in the electric field is stored in the form of potential energy of g.
If V be the electric potential at the present location of the point charge g, then the work done in bringing it from infinity to the present location is qV. The potential energy of q at r in an external field is U q. r ….*potential energy+
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where V r is the potential at r.
(ii)
Potential energy of a system of two charges: Suppose two charges q1 and q2 are located at r1 and r2 respectively in an external field. To find the potential energy of the two charge system, first bring q1 from infinity to r1. The work done in this process is
in
W1 = q1 V( r1 ), where V( r1 ) is the potential at r1 due to the external field E .
e.
Next, we bring q2 from infinity to r2. W2 = q2 V( r2 ),
iit je
The work done on q2 against the external field E is
cr ac k
where V( r2 ) is the potential at r2 due to the external field E only. The work done on q2 against the field of q1 is W3 =
q1q2 , 40r12
where r12 is the separation between q1 and q2. The potential energy of the two-charge system is U = W 1 + W2 + W3 = qV r1 q2V r2
q1q2 40r12
Example : Two protons are separated by a distance Ft. What will be the speed of each proton when they reach infinity under their mutual repulsion?
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Solution: Let the charge on each proton be Q. Initially they are at rest, so their kinetic energy (KE) = 0 and they have only potential energy equal to
Q2 . 40R Finally at infinity, they have infinite separation so their potential energy will become zero. (because R =
1 2
1 mv2 = mv2 (where m is the mass of each 2
.in
mv2 for each proton) 2x
now.) And total kinetic energy is, (
v=
cr a
Therefore,
itj
Q2 mv2 40R
ck i
or,
ee
proton and v is the speed with which they are moving at infinity). Now, total initial energy = total final energy
Q . 40mR
Example: A bullet of mass 2 gm is moving with a speed of 10 m/s. If the bullet has a charge of 2 micro-coulomb, through what potential it be accelerated starting from rest, to acquire the same speed? Solution: Use the relation
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qV =
1 2 mv ; 2
Here, q = 2 × 10–6 coulomb; m = 2 × 10–3 kg; v = 10 m/s Therefore,
mv2 2q
101 = 2 106 = 5 × 104 volt
Example :
in
cr ac k
= 50 kV
2
e.
2 103 10 = 2 2 106
iit je
V=
Three equal charges Q are at the vertices of an equilateral triangle of side A. How much work is done in bringing them closer to an equilateral triangle of side
A ? 2
Solution: In this problem first we have to calculate the initial and final potential energy and then the work done is final potential energy minus initial potential energy.
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1 3Q2 Initial potential energy (for each combination, the energy is 40 A Q2 and we have three such combinations) 40A Final potential energy
A 2
in
40
Now, work done
Therefore,
Example :
cr ac k
3Q2 W= 40 A
iit je
3Q2 3Q2 W= A 40 A 40 2
e.
=
3Q2
Two particles have equal masses of 5 g each and opposite charges of 4 × 1 0–5 C and – 4 × 10–5 C. They are released from rest with a separation of 1 m between them. Find the speed of particles when the separation is reduced to 50 cm. Solution: Again in this question we have to do the conservation of mechanical energy. Initially, there is only potential energy but finally there is both kinetic and potential energy. Initial state energy:
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Bodies are at rest so kinetic energy = 0
Q1Q2 40R
Potential energy =
=
9 109 4 105 4 105
1
= – 14.4 joule
in
Final state energy:
iit je
e.
Let us suppose both the charges approach each other with individual speed v. (both will have same speed because masses are same and momentum will be conserved. Since initial momentum is zero, they will have final momentum also equal to zero. Since momentum is vector, they will have equal and opposite momentum)
cr ac k
Final kinetic energy of both =2
1 mv2 2
= mv2
Final potential energy =
=
Q1Q2 40R
9 109 4 105 4 105
0.5
= – 28.8 joule Now,
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initial total energy = final total energy – 14.4 = mv2 – 28.8 therefore, mv2 = 14.4; or, v =
14.4 5 103
= 54 m/s
in
Work done in rotating an Electric Dipole in an Electric Field:
iit je
e.
If a dipole, placed in a uniform electric field, is rotated from its equilibrium position, work has to be done against the electric field torque.
cr ac k
If a small work dW is done (against the field torque) in rotating the dipole through a small angle dӨ, then dW = Torque × angular displacement or dW = Ө × dӨ = pE sin Ө dӨ Thus, the total work done against the field torque for a deflection ' ' from equilibrium is
W = pE sin d 0
= pE cos 0
W = pE(1 – cos Ө) This is the formula for the work done in rotating an electric dipole against an electric field through an angleӨ from the direction of the field (equilibrium position). The work done in rotating from the position 1 to Ө2 is given by W = pE(cos Ө1 – cos Ө2).
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Case (i) If the dipole be rotated through 90° from the direction of the field, then the work done: W = pE(1 – cos 90°) = pE (1 – 0) = pE.
Case (ii)
e.
in
The work done in rotating the dipole through 180° from the direction of the field
= pE [1 – (–1)] = 2pE.
cr ac k
NOTE:
iit je
W= pE (1 – cos 180°)
The work done by the field torque on the dipole is W = pE (cos Ө2 – cos Ө1) = – pE(1 – cos Ө),
when Ө1 = 0°, Ө2 = Ө.
Potential Energy of an Electric Dipole in an Electric Field. The potential energy of an electric dipole in an electric field is defined as the work done in bringing the dipole from infinity to inside the field. An electric dipole (+q, –q) is brought from infinity to a uniform electric field E in such a way that the dipole moment p is always in the direction
of the field. Due to the field E , a force F qE acts on the charge +q
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in the direction of the field, and force F qE on the charge –q in the opposite direction. Hence, in bringing the dipole in the field, work will be done on the charge +q by an external agent, while work will be done by the field itself on the charge –q. But, as the dipole is brought from infinity into the field, the charge –q covers 2l distance more than the charge +q. Therefore, the work done on –q will be greater. Hence the 'net' work done in bringing the dipole from infinity into the field = force on charge (–q) × additional distance moved = – qE × 2l
in
= – pE.
iit je
e.
This work is the potential energy U0 of the electric dipole placed in the electric field parallel to it: U0 = – pE
cr ac k
In this position the electric dipole is in stable equilibrium inside the field.
On rotating the dipole through an angle Ө work will have to be done on the dipole. This work is given by W = pE(1 – cos Ө).
This will result in an increase in the potential energy of the dipole. Hence, the potential energy of the dipole in the position Ө will be given by UӨ = U0 + W = – pE + pE (1 – cos Ө). or UӨ = – pE cos Ө or, U p.E This is the general equation of the potential energy of the electric dipole.
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DIPOLE IN A UNIFORM EXTERNAL FIELD Consider a small dipole (+q, –q) of length 2l, placed in uniform electric field of intensity E . Forces of magnitude qE each act on the charge +q, and –q as shown in figure. These forces are equal, parallel and opposite, and hence net force on the dipole is zero but these forces constitute a couple. The moment of this couple i.e., torque is Ʈ = force × perpendicular distance
in
= F × 2l sin Ө = qE 2l sin Ө
e.
= 2ql E sin Ө
iit je
or = pE sin Ө,
where 2ql = p (electric dipole moment).
cr ac k
Ʈ= pE sin Ө
or, p E
If the dipole is placed perpendicular to the electric field (Ө= 90° or sin Ө = 1), then the couple acting on it will be maximum. If this be Өmax, then Өmax = PE. If E = 1 NC–1, then p = Өmax. Hence, electric dipole moment is the torque acting on the dipole placed perpendicular to the direction of a uniform electric field of intensity 1 NC–1.
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Period of Dipole in Uniform Electric Field 1.
If dipole is parallel or antiparallel to the field (i.e., = 0° or 180°), torque is zero. Hence dipole is in equilibrium position. Consider an electric dipole of dipole moment p making an angle 'Ө' with the field direction. It experiences a torque pE sin tending to bring it in the direction of field. Therefore, on being released, the dipole oscillates about an axis through its centre of mass and perpendicular to the field. If l is the moment of inertia of the dipole about the axis of rotation, then the
d2 = – pE sin Ө dt2
e.
l
in
equation of motion is:
iit je
For small amplitudes sin Ө - Ө.
d2 pE Thus, dt2 l
cr ac k
= – Ѡ2Ө,
where
pE l
This is an SHM whose period of oscillation is T=
2
= 2 2.
l . pE
Potential energy of the electric dipole at an angle Ө with electric field is U P.E = PE cos Ө.
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3.
Work done is rotating the dipole from angle Ө1 to Ө2 is W = PE (cos Ө1 – cosӨ2).
Example : An electric dipole consists of two opposite charges of magnitude 0.2 µC each, separated by a distance of 2 cm. The dipole is placed in an external field of 2 × 10s N/C. What maximum torque does the field exert on the dipole? Solution:
in
Torque = pE sin Ө
iit je
maximum torque = pE
e.
For maximum value of torque, sin = 1 = (0.2 × 10–6 × 2 × 10–2) (2 × 105) = 8 × 10–4 Nm
cr ac k
Motion of a Charged Particle In a Uniform Electric Field Along x-axis, ux = v0, ax = 0, Sx = x,
vx = v0 S = ut
1 2 at 2
x = v 0t Along y-axis, uy = 0,
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ay = y=
qE , m 1 qE 2 t 2 m
1 qE x y= 2 m v0
2
1 qEx2 y 2 mv20
in
Ө y Ө x2
iit je
e.
i.e., path is parabolic.
Deflection on the screen Y = (y + D tan Ө)
vy vx
qEt qEx mv0 mv20
cr ac k
tan Ө =
qEx2 2 So tan Ө= 2mv20 x tan
y x 2
which means that the tangent at point P intersect the line AB at its mid point i.e. at distance
x from B. 2
x Y D tan 2 (i)
Electric Field at a point on the Axis of a Dipole (end-on position).
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Suppose an electric dipole (q, –q) of length 2l is situated in a medium of dielectric constant K. Let P be a point on the axis at a distance r metre from the mid-point O of the dipole. We have to determine the intensity of the electric field at P. (In figure)
Let E and E2 be the intensities of the electric fields at P due to the charges +q and –q of the dipole, respectively. The distance of the point P from the charge +q is (r – l) and the distance from charge – q is (r + l). Therefore,
(away from charge +q)
1 q 40K r l2
iit je
and E2 =
in
1 q 40K r l2
e.
E1 =
cr ac k
(towards the charge –q)
The fields E1 and E, are along the same line but in opposite directions. Therefore, the resultant field E at the point P will be E = E1 – E2 =
1 q 1 q 2 40K r l 40K r l2
1 1 q = 40K r l2 r l2 directed from –q to +q =
4q r
40K r2 l2
2
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or E =
2pr
40K r2 l2
2
directed from –q to +q. If l is very small compared to r(l > l. Therefore,
1 p 40 r3
iit je
or E
in
1 p 40 r3
e.
E=
cr ac k
The direction of the electric field is parallel to the axis of the dipole.
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Electric Field due to a Dipole at any Point for a Small Dipole Suppose the point P on the line making an angle with the axis of a small dipole (l
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