DC Pandey Solutions: Electrostatics - 2 Notes | Study Physics Class 12 - NEET

NEET: DC Pandey Solutions: Electrostatics - 2 Notes | Study Physics Class 12 - NEET

The document DC Pandey Solutions: Electrostatics - 2 Notes | Study Physics Class 12 - NEET is a part of the NEET Course Physics Class 12.
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 Page 1


Introductory Exercise 21.4 
Q 1.  A point charge q
1
 =1.0 ?C is held fixed at origin. A second point charge q
2
 =-2.0 ?C and a mass 
10
-4
 kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from 
rest. What is its speed when it is 0.5 m from the origin? 
Q 2.  A point charge q
1
 =-1.0 ?C is held stationary at the origin. A second point charge q
2
 = + 2.0 ?C 
moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force 
on q
2
? 
Q 3.  A point charge q
1
 is held stationary at the origin. A second charge q
2
 is placed at a point a, and the 
electric potential energy of the pair of charges is -6.4 ×10
-8
 J. When the second charge is moved to 
point b, the electric force on the charge does 4.2 × 10
-8
 J of work. What is the electric potential 
energy of the pair of charges when the second charge is at point b ? 
Q 4.  Is it possible to have an arrangement of two point charges separated by finite distances such that 
the electric potential energy of the arrangement is the same as if the two charges were infinitely far 
apart? What if there are three charges? 
Solutions 
1.  K
1
 +U
1 
= K
f
 + U
f
  
  ?  
2 1 2 1 2
0 1 0 f
q q q q 1 1 1
mv
4 r 2 4 r
??
? ? ? ?
 
  ?  
12
0 i f
2 1 1 1
v (q q )
m 4 r r
?? ??
??
?? ??
??
?? ??
 
   
9 12
4
2 1 1
(9 10 )( 2 10 )
10 1.0 0.5
?
?
??
? ? ? ? ?
??
??
 
   = 18.97 m/s  
2.  Work done by electrostatic forces 
  = - ?U 
   =U
i
 - U
f
  
  
12
0 i f
1 1 1
(q q )
4 r r
?? ??
??
?? ??
??
?? ??
 
   = (9 × 10
9
)(-2 × 10
-12
) 
11
0.1 2.0
??
?
??
??
  
  = -9 × 10
-3
J = -9mJ  
3.  Work done by electrostatic forces 
   W = - ?U = U
i
 - U
f
 
? U
0 
= U
t 
- W 
  = (- 6.4 × 10
-8
) - (4.2 × 10
-8
) = - 10.6 × 10
-8
 J 
4. U
?
 = 0 
  
12
r
0
qq 1
U.
4r
?
??
  
  (For two charges)  
Page 2


Introductory Exercise 21.4 
Q 1.  A point charge q
1
 =1.0 ?C is held fixed at origin. A second point charge q
2
 =-2.0 ?C and a mass 
10
-4
 kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from 
rest. What is its speed when it is 0.5 m from the origin? 
Q 2.  A point charge q
1
 =-1.0 ?C is held stationary at the origin. A second point charge q
2
 = + 2.0 ?C 
moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force 
on q
2
? 
Q 3.  A point charge q
1
 is held stationary at the origin. A second charge q
2
 is placed at a point a, and the 
electric potential energy of the pair of charges is -6.4 ×10
-8
 J. When the second charge is moved to 
point b, the electric force on the charge does 4.2 × 10
-8
 J of work. What is the electric potential 
energy of the pair of charges when the second charge is at point b ? 
Q 4.  Is it possible to have an arrangement of two point charges separated by finite distances such that 
the electric potential energy of the arrangement is the same as if the two charges were infinitely far 
apart? What if there are three charges? 
Solutions 
1.  K
1
 +U
1 
= K
f
 + U
f
  
  ?  
2 1 2 1 2
0 1 0 f
q q q q 1 1 1
mv
4 r 2 4 r
??
? ? ? ?
 
  ?  
12
0 i f
2 1 1 1
v (q q )
m 4 r r
?? ??
??
?? ??
??
?? ??
 
   
9 12
4
2 1 1
(9 10 )( 2 10 )
10 1.0 0.5
?
?
??
? ? ? ? ?
??
??
 
   = 18.97 m/s  
2.  Work done by electrostatic forces 
  = - ?U 
   =U
i
 - U
f
  
  
12
0 i f
1 1 1
(q q )
4 r r
?? ??
??
?? ??
??
?? ??
 
   = (9 × 10
9
)(-2 × 10
-12
) 
11
0.1 2.0
??
?
??
??
  
  = -9 × 10
-3
J = -9mJ  
3.  Work done by electrostatic forces 
   W = - ?U = U
i
 - U
f
 
? U
0 
= U
t 
- W 
  = (- 6.4 × 10
-8
) - (4.2 × 10
-8
) = - 10.6 × 10
-8
 J 
4. U
?
 = 0 
  
12
r
0
qq 1
U.
4r
?
??
  
  (For two charges)  
  U
r
 ? U
?
 
   for r ? ? 
  
2 3 3 1 12
r
0 12 23 31
q q q q qq 1
U
4 r r r
??
? ? ?
??
??
??
 
  Now U
r
 can be equal to U
?
 for finite value of r. 
Introductory Exercise 21.5 
Q 1.  Find V
ba
 if 12 J of work has to be done against an electric field to take a charge of 10
-2
 C from a to 
b. 
Q 2.  A rod of length L lies along the x-axis with its left end at the origin. It has a nonuniform charge 
density ? = ?x, where ? is a positive constant. 
  (a) What are the units of ??  
  (b) Calculate the electric potential at point A where x = - d. 
Q 3.  A charge q is uniformly distributed along an insulating straight wire of length 2/ as shown in Fig. 
Find an expression for the electric potential at a point located a distance c/from the distribution 
along its perpendicular bisector. 
 
Q 4.  A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. 
Calculate the work done in bringing a small test charge q from infinity to the apex of the cone. The 
cone has a slope length L. 
Solutions 
1.  W
ab 
= ?U
a-b
 = U
b 
- U
a 
= q(V
b 
- V
a
) = qV
ba
 
  ?  
ab
ba 2
W 12
V
q 10
?
?? 
   = 1200 V 
2.  (a) 
2
C / m C
x m m
?
? ? ? ? 
  (b) 
  
0
1 dq
dV .
4 x d
?
? ? ?
 
   
0
1 dx
4 x d
?? ? ??
?
????
? ? ?
??
??
 
   
0
1 x dx
4 x d
?? ???
?
????
? ? ?
??
??
 
Page 3


Introductory Exercise 21.4 
Q 1.  A point charge q
1
 =1.0 ?C is held fixed at origin. A second point charge q
2
 =-2.0 ?C and a mass 
10
-4
 kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from 
rest. What is its speed when it is 0.5 m from the origin? 
Q 2.  A point charge q
1
 =-1.0 ?C is held stationary at the origin. A second point charge q
2
 = + 2.0 ?C 
moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force 
on q
2
? 
Q 3.  A point charge q
1
 is held stationary at the origin. A second charge q
2
 is placed at a point a, and the 
electric potential energy of the pair of charges is -6.4 ×10
-8
 J. When the second charge is moved to 
point b, the electric force on the charge does 4.2 × 10
-8
 J of work. What is the electric potential 
energy of the pair of charges when the second charge is at point b ? 
Q 4.  Is it possible to have an arrangement of two point charges separated by finite distances such that 
the electric potential energy of the arrangement is the same as if the two charges were infinitely far 
apart? What if there are three charges? 
Solutions 
1.  K
1
 +U
1 
= K
f
 + U
f
  
  ?  
2 1 2 1 2
0 1 0 f
q q q q 1 1 1
mv
4 r 2 4 r
??
? ? ? ?
 
  ?  
12
0 i f
2 1 1 1
v (q q )
m 4 r r
?? ??
??
?? ??
??
?? ??
 
   
9 12
4
2 1 1
(9 10 )( 2 10 )
10 1.0 0.5
?
?
??
? ? ? ? ?
??
??
 
   = 18.97 m/s  
2.  Work done by electrostatic forces 
  = - ?U 
   =U
i
 - U
f
  
  
12
0 i f
1 1 1
(q q )
4 r r
?? ??
??
?? ??
??
?? ??
 
   = (9 × 10
9
)(-2 × 10
-12
) 
11
0.1 2.0
??
?
??
??
  
  = -9 × 10
-3
J = -9mJ  
3.  Work done by electrostatic forces 
   W = - ?U = U
i
 - U
f
 
? U
0 
= U
t 
- W 
  = (- 6.4 × 10
-8
) - (4.2 × 10
-8
) = - 10.6 × 10
-8
 J 
4. U
?
 = 0 
  
12
r
0
qq 1
U.
4r
?
??
  
  (For two charges)  
  U
r
 ? U
?
 
   for r ? ? 
  
2 3 3 1 12
r
0 12 23 31
q q q q qq 1
U
4 r r r
??
? ? ?
??
??
??
 
  Now U
r
 can be equal to U
?
 for finite value of r. 
Introductory Exercise 21.5 
Q 1.  Find V
ba
 if 12 J of work has to be done against an electric field to take a charge of 10
-2
 C from a to 
b. 
Q 2.  A rod of length L lies along the x-axis with its left end at the origin. It has a nonuniform charge 
density ? = ?x, where ? is a positive constant. 
  (a) What are the units of ??  
  (b) Calculate the electric potential at point A where x = - d. 
Q 3.  A charge q is uniformly distributed along an insulating straight wire of length 2/ as shown in Fig. 
Find an expression for the electric potential at a point located a distance c/from the distribution 
along its perpendicular bisector. 
 
Q 4.  A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. 
Calculate the work done in bringing a small test charge q from infinity to the apex of the cone. The 
cone has a slope length L. 
Solutions 
1.  W
ab 
= ?U
a-b
 = U
b 
- U
a 
= q(V
b 
- V
a
) = qV
ba
 
  ?  
ab
ba 2
W 12
V
q 10
?
?? 
   = 1200 V 
2.  (a) 
2
C / m C
x m m
?
? ? ? ? 
  (b) 
  
0
1 dq
dV .
4 x d
?
? ? ?
 
   
0
1 dx
4 x d
?? ? ??
?
????
? ? ?
??
??
 
   
0
1 x dx
4 x d
?? ???
?
????
? ? ?
??
??
 
  ?  
xL
x0
0
L
V dV L dln 1
4d
?
?
? ? ? ??
? ? ? ?
?? ??
??
?? ??
?
 
3.  
   
q
dq dx
2l
??
?
??
??
 
  At point P
 
 
   
0
1 dq
dV
4r
??
??
?
????
??
??
??
 
    
22
0
q
.dx
1
2l
4
dx
??
?? ??
?
?? ??
??
? ??
??
??
 
  ?  
xl
x0
V 2 dV
?
?
?
?
 
4.  
  W = ?U = U
apex 
- U
?
 
  W = qV
apex
 ...(i) 
  As U
?
 = 0  
  Vapex 
   
rR
xL
? 
  ?  
R
rx
L
??
?
??
??
 
  Surface charge density 
   
Q
Rl
??
?
 
  dq = ( ?) (dA)  
    
Q
(2 r)dx
Rl
??
??
??
?
??
 
   
R
(2 ) x dx
RL L
? ? ? ? ?
??
? ? ? ?
?
? ? ? ?
 
Page 4


Introductory Exercise 21.4 
Q 1.  A point charge q
1
 =1.0 ?C is held fixed at origin. A second point charge q
2
 =-2.0 ?C and a mass 
10
-4
 kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from 
rest. What is its speed when it is 0.5 m from the origin? 
Q 2.  A point charge q
1
 =-1.0 ?C is held stationary at the origin. A second point charge q
2
 = + 2.0 ?C 
moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force 
on q
2
? 
Q 3.  A point charge q
1
 is held stationary at the origin. A second charge q
2
 is placed at a point a, and the 
electric potential energy of the pair of charges is -6.4 ×10
-8
 J. When the second charge is moved to 
point b, the electric force on the charge does 4.2 × 10
-8
 J of work. What is the electric potential 
energy of the pair of charges when the second charge is at point b ? 
Q 4.  Is it possible to have an arrangement of two point charges separated by finite distances such that 
the electric potential energy of the arrangement is the same as if the two charges were infinitely far 
apart? What if there are three charges? 
Solutions 
1.  K
1
 +U
1 
= K
f
 + U
f
  
  ?  
2 1 2 1 2
0 1 0 f
q q q q 1 1 1
mv
4 r 2 4 r
??
? ? ? ?
 
  ?  
12
0 i f
2 1 1 1
v (q q )
m 4 r r
?? ??
??
?? ??
??
?? ??
 
   
9 12
4
2 1 1
(9 10 )( 2 10 )
10 1.0 0.5
?
?
??
? ? ? ? ?
??
??
 
   = 18.97 m/s  
2.  Work done by electrostatic forces 
  = - ?U 
   =U
i
 - U
f
  
  
12
0 i f
1 1 1
(q q )
4 r r
?? ??
??
?? ??
??
?? ??
 
   = (9 × 10
9
)(-2 × 10
-12
) 
11
0.1 2.0
??
?
??
??
  
  = -9 × 10
-3
J = -9mJ  
3.  Work done by electrostatic forces 
   W = - ?U = U
i
 - U
f
 
? U
0 
= U
t 
- W 
  = (- 6.4 × 10
-8
) - (4.2 × 10
-8
) = - 10.6 × 10
-8
 J 
4. U
?
 = 0 
  
12
r
0
qq 1
U.
4r
?
??
  
  (For two charges)  
  U
r
 ? U
?
 
   for r ? ? 
  
2 3 3 1 12
r
0 12 23 31
q q q q qq 1
U
4 r r r
??
? ? ?
??
??
??
 
  Now U
r
 can be equal to U
?
 for finite value of r. 
Introductory Exercise 21.5 
Q 1.  Find V
ba
 if 12 J of work has to be done against an electric field to take a charge of 10
-2
 C from a to 
b. 
Q 2.  A rod of length L lies along the x-axis with its left end at the origin. It has a nonuniform charge 
density ? = ?x, where ? is a positive constant. 
  (a) What are the units of ??  
  (b) Calculate the electric potential at point A where x = - d. 
Q 3.  A charge q is uniformly distributed along an insulating straight wire of length 2/ as shown in Fig. 
Find an expression for the electric potential at a point located a distance c/from the distribution 
along its perpendicular bisector. 
 
Q 4.  A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. 
Calculate the work done in bringing a small test charge q from infinity to the apex of the cone. The 
cone has a slope length L. 
Solutions 
1.  W
ab 
= ?U
a-b
 = U
b 
- U
a 
= q(V
b 
- V
a
) = qV
ba
 
  ?  
ab
ba 2
W 12
V
q 10
?
?? 
   = 1200 V 
2.  (a) 
2
C / m C
x m m
?
? ? ? ? 
  (b) 
  
0
1 dq
dV .
4 x d
?
? ? ?
 
   
0
1 dx
4 x d
?? ? ??
?
????
? ? ?
??
??
 
   
0
1 x dx
4 x d
?? ???
?
????
? ? ?
??
??
 
  ?  
xL
x0
0
L
V dV L dln 1
4d
?
?
? ? ? ??
? ? ? ?
?? ??
??
?? ??
?
 
3.  
   
q
dq dx
2l
??
?
??
??
 
  At point P
 
 
   
0
1 dq
dV
4r
??
??
?
????
??
??
??
 
    
22
0
q
.dx
1
2l
4
dx
??
?? ??
?
?? ??
??
? ??
??
??
 
  ?  
xl
x0
V 2 dV
?
?
?
?
 
4.  
  W = ?U = U
apex 
- U
?
 
  W = qV
apex
 ...(i) 
  As U
?
 = 0  
  Vapex 
   
rR
xL
? 
  ?  
R
rx
L
??
?
??
??
 
  Surface charge density 
   
Q
Rl
??
?
 
  dq = ( ?) (dA)  
    
Q
(2 r)dx
Rl
??
??
??
?
??
 
   
R
(2 ) x dx
RL L
? ? ? ? ?
??
? ? ? ?
?
? ? ? ?
 
   
2
2Q
x dx
L
??
?
??
??
 
 Now 
0
1 dq
dV
4x
??
??
?
????
??
??
??
 
   
2
0
2Q
dx
4L
??
?
??
??
??
 
  ?  
L
0
V dV ?
?
 
   
0
Q
2L
?
??
 
 Substituting in Eq. (i), we have 
   
0
Qq
W
2L
?
??
 
Introductory Exercise 21.6 
Q 1.  Determine the electric field strength vector if the potential of this field depends on x, y co-
ordinates as 
  (a) V = a(x
2 
- y
2
)  (b) V = axy  
  where, a is a constant. 
Q 2.  The electrical potential function for an electrical field directed parallel to the x-axis is shown in the 
given graph. 
 
  Draw the graph of electric field strength.  
Q 3.  The electric potential decreases uniformly from 100 V to 50 V as one moves along the x-axis from 
x = 0 to x = 5 m. The electric field at x = 2 m must be equal to 10 V/m. Is this statement true or 
false. 
Q 4.  In the uniform electric field shown in figure, find :  
 
  (a) V
A 
- V
D  
(b) V
A 
- V
C
   (c)V
B 
- V
D  
(d)V
C 
- V
D  
Solutions 
1.  (a)  
VV
ˆˆ
E i j
xy
?? ??
? ? ?
??
??
??
?
    
  
ˆ ˆ ˆ ˆ
a[(2x)i (2y)j] 2a[xi yj] ? ? ? ? ? ? 
Page 5


Introductory Exercise 21.4 
Q 1.  A point charge q
1
 =1.0 ?C is held fixed at origin. A second point charge q
2
 =-2.0 ?C and a mass 
10
-4
 kg is placed on the x-axis, 1.0 m from the origin. The second point charge is released from 
rest. What is its speed when it is 0.5 m from the origin? 
Q 2.  A point charge q
1
 =-1.0 ?C is held stationary at the origin. A second point charge q
2
 = + 2.0 ?C 
moves from the point (1.0 m, 0, 0) to (2.0 m, 0, 0). How much work is done by the electric force 
on q
2
? 
Q 3.  A point charge q
1
 is held stationary at the origin. A second charge q
2
 is placed at a point a, and the 
electric potential energy of the pair of charges is -6.4 ×10
-8
 J. When the second charge is moved to 
point b, the electric force on the charge does 4.2 × 10
-8
 J of work. What is the electric potential 
energy of the pair of charges when the second charge is at point b ? 
Q 4.  Is it possible to have an arrangement of two point charges separated by finite distances such that 
the electric potential energy of the arrangement is the same as if the two charges were infinitely far 
apart? What if there are three charges? 
Solutions 
1.  K
1
 +U
1 
= K
f
 + U
f
  
  ?  
2 1 2 1 2
0 1 0 f
q q q q 1 1 1
mv
4 r 2 4 r
??
? ? ? ?
 
  ?  
12
0 i f
2 1 1 1
v (q q )
m 4 r r
?? ??
??
?? ??
??
?? ??
 
   
9 12
4
2 1 1
(9 10 )( 2 10 )
10 1.0 0.5
?
?
??
? ? ? ? ?
??
??
 
   = 18.97 m/s  
2.  Work done by electrostatic forces 
  = - ?U 
   =U
i
 - U
f
  
  
12
0 i f
1 1 1
(q q )
4 r r
?? ??
??
?? ??
??
?? ??
 
   = (9 × 10
9
)(-2 × 10
-12
) 
11
0.1 2.0
??
?
??
??
  
  = -9 × 10
-3
J = -9mJ  
3.  Work done by electrostatic forces 
   W = - ?U = U
i
 - U
f
 
? U
0 
= U
t 
- W 
  = (- 6.4 × 10
-8
) - (4.2 × 10
-8
) = - 10.6 × 10
-8
 J 
4. U
?
 = 0 
  
12
r
0
qq 1
U.
4r
?
??
  
  (For two charges)  
  U
r
 ? U
?
 
   for r ? ? 
  
2 3 3 1 12
r
0 12 23 31
q q q q qq 1
U
4 r r r
??
? ? ?
??
??
??
 
  Now U
r
 can be equal to U
?
 for finite value of r. 
Introductory Exercise 21.5 
Q 1.  Find V
ba
 if 12 J of work has to be done against an electric field to take a charge of 10
-2
 C from a to 
b. 
Q 2.  A rod of length L lies along the x-axis with its left end at the origin. It has a nonuniform charge 
density ? = ?x, where ? is a positive constant. 
  (a) What are the units of ??  
  (b) Calculate the electric potential at point A where x = - d. 
Q 3.  A charge q is uniformly distributed along an insulating straight wire of length 2/ as shown in Fig. 
Find an expression for the electric potential at a point located a distance c/from the distribution 
along its perpendicular bisector. 
 
Q 4.  A cone made of insulating material has a total charge Q spread uniformly over its sloping surface. 
Calculate the work done in bringing a small test charge q from infinity to the apex of the cone. The 
cone has a slope length L. 
Solutions 
1.  W
ab 
= ?U
a-b
 = U
b 
- U
a 
= q(V
b 
- V
a
) = qV
ba
 
  ?  
ab
ba 2
W 12
V
q 10
?
?? 
   = 1200 V 
2.  (a) 
2
C / m C
x m m
?
? ? ? ? 
  (b) 
  
0
1 dq
dV .
4 x d
?
? ? ?
 
   
0
1 dx
4 x d
?? ? ??
?
????
? ? ?
??
??
 
   
0
1 x dx
4 x d
?? ???
?
????
? ? ?
??
??
 
  ?  
xL
x0
0
L
V dV L dln 1
4d
?
?
? ? ? ??
? ? ? ?
?? ??
??
?? ??
?
 
3.  
   
q
dq dx
2l
??
?
??
??
 
  At point P
 
 
   
0
1 dq
dV
4r
??
??
?
????
??
??
??
 
    
22
0
q
.dx
1
2l
4
dx
??
?? ??
?
?? ??
??
? ??
??
??
 
  ?  
xl
x0
V 2 dV
?
?
?
?
 
4.  
  W = ?U = U
apex 
- U
?
 
  W = qV
apex
 ...(i) 
  As U
?
 = 0  
  Vapex 
   
rR
xL
? 
  ?  
R
rx
L
??
?
??
??
 
  Surface charge density 
   
Q
Rl
??
?
 
  dq = ( ?) (dA)  
    
Q
(2 r)dx
Rl
??
??
??
?
??
 
   
R
(2 ) x dx
RL L
? ? ? ? ?
??
? ? ? ?
?
? ? ? ?
 
   
2
2Q
x dx
L
??
?
??
??
 
 Now 
0
1 dq
dV
4x
??
??
?
????
??
??
??
 
   
2
0
2Q
dx
4L
??
?
??
??
??
 
  ?  
L
0
V dV ?
?
 
   
0
Q
2L
?
??
 
 Substituting in Eq. (i), we have 
   
0
Qq
W
2L
?
??
 
Introductory Exercise 21.6 
Q 1.  Determine the electric field strength vector if the potential of this field depends on x, y co-
ordinates as 
  (a) V = a(x
2 
- y
2
)  (b) V = axy  
  where, a is a constant. 
Q 2.  The electrical potential function for an electrical field directed parallel to the x-axis is shown in the 
given graph. 
 
  Draw the graph of electric field strength.  
Q 3.  The electric potential decreases uniformly from 100 V to 50 V as one moves along the x-axis from 
x = 0 to x = 5 m. The electric field at x = 2 m must be equal to 10 V/m. Is this statement true or 
false. 
Q 4.  In the uniform electric field shown in figure, find :  
 
  (a) V
A 
- V
D  
(b) V
A 
- V
C
   (c)V
B 
- V
D  
(d)V
C 
- V
D  
Solutions 
1.  (a)  
VV
ˆˆ
E i j
xy
?? ??
? ? ?
??
??
??
?
    
  
ˆ ˆ ˆ ˆ
a[(2x)i (2y)j] 2a[xi yj] ? ? ? ? ? ? 
  (b) Again 
   
VV
ˆˆ
E i j
xy
?? ??
? ? ?
??
??
??
?
 
   
ˆˆ
[yi xj] ? ? ? ?  
2.  
dV
E
dx
?? = - Slope of V - x graph. 
  From x = - 2m to x = 0, slope = + 5 V/m 
  E = -5 V/m From x = 0 to x = 2m, slope = 0, 
  E = 0 From x = 2m to x = 4m, slope = + 5 V/m, 
  E =-5 V/m From x = 4m to x = 8m slope = - 5 V/m 
  E = + 5 V/m Corresponding E-x graph is as shown in answer. 
3. 
V 50
x5
??
?
?
= -10 V/m 
  Now, 
2
22
VVV
| E |
x y z
?? ??? ? ? ? ?
? ? ?
?? ? ? ? ?
? ? ?
? ? ? ?
??
?
 
  No information is given about 
V
y
?
?
and 
V
z
?
?
.
  
  
Hence 
   
V
| E |
x
?
?
?
?
 
  or | E | ?
?
10 V/m 
4.  V
A
 = V
D
 and V
B
 = V
C
 as the points A and D or B and C are lying on same equipotential surface ( ?  
to electric field lines). Further, V
A
 or V
D
 > V
B
 or V
C
 as electric lines always flow from higher 
potential to lower potential 
  V
A 
- V
B
 = V
D 
- V
C
 = Ed 
   = (20) (1) 
  = 20 V  
Introductory Exercise 21.7 
Q 1.  Show that the torque on an electric dipole placed in a uniform electric field is, 
   PE ? ? ?
??
?
 
  independent of the origin about which torque is calculated. 
Q 2.  Three point charges q, - 2q and q are located along the x-axis as shown in figure. Show that the 
electric field at P(y >> a) along the y-axis is, 
   
2
4
0
1 3qa
ˆ
Ej
4y
??
??
?
 
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