Electric Charges and Fields Practice Questions - DPP for JEE

 Page 1


1. (a) Surface charge density (s) = 
So 
and 
2. (d) Since electric field  decreases inside water, therefore flux 
 also decreases.
3. (c) When a dipole is placed in a uniform electric field, two equal and
opposite forces act on it. Therefore, a torque acts which rotates the
dipole.
4. (a) For the distances close to the charge at x = 0 the field is very high
and is in positive direction of x-axis. As we move towards the
other charge the net electric field becomes zero at x = a thereafter
the influence of charge at x = 2a dominates and net field increases
in negative direction of x-axis and grows unboundedly as we come
closer and closer to the charge at x = 2a.
5. (b) From figure, , where
Page 2


1. (a) Surface charge density (s) = 
So 
and 
2. (d) Since electric field  decreases inside water, therefore flux 
 also decreases.
3. (c) When a dipole is placed in a uniform electric field, two equal and
opposite forces act on it. Therefore, a torque acts which rotates the
dipole.
4. (a) For the distances close to the charge at x = 0 the field is very high
and is in positive direction of x-axis. As we move towards the
other charge the net electric field becomes zero at x = a thereafter
the influence of charge at x = 2a dominates and net field increases
in negative direction of x-axis and grows unboundedly as we come
closer and closer to the charge at x = 2a.
5. (b) From figure, , where
or
or
or
6. (b) Let us consider a spherical shell of radius x and thickness dx.
Charge on this shell
dq = ?.4px
2
dx = 
? Total charge in the spherical region from centre to r (r < R) is
Page 3


1. (a) Surface charge density (s) = 
So 
and 
2. (d) Since electric field  decreases inside water, therefore flux 
 also decreases.
3. (c) When a dipole is placed in a uniform electric field, two equal and
opposite forces act on it. Therefore, a torque acts which rotates the
dipole.
4. (a) For the distances close to the charge at x = 0 the field is very high
and is in positive direction of x-axis. As we move towards the
other charge the net electric field becomes zero at x = a thereafter
the influence of charge at x = 2a dominates and net field increases
in negative direction of x-axis and grows unboundedly as we come
closer and closer to the charge at x = 2a.
5. (b) From figure, , where
or
or
or
6. (b) Let us consider a spherical shell of radius x and thickness dx.
Charge on this shell
dq = ?.4px
2
dx = 
? Total charge in the spherical region from centre to r (r < R) is
= 
= 
= 
? Electric field at r,  E = 
= 
= 
7. (a) The flux is zero according to Gauss’ Law because it is a open
surface which enclosed a charge q.
8. (b)
Electric field at C due to electric dipole
=  along OC
Electric field at C due to induced charge must be equal and opposite to
electric field due to dipole as net field at C is zero.
9. (a) ? = linear charge density;
Charge on elementary portion dx = ? dx.
Page 4


1. (a) Surface charge density (s) = 
So 
and 
2. (d) Since electric field  decreases inside water, therefore flux 
 also decreases.
3. (c) When a dipole is placed in a uniform electric field, two equal and
opposite forces act on it. Therefore, a torque acts which rotates the
dipole.
4. (a) For the distances close to the charge at x = 0 the field is very high
and is in positive direction of x-axis. As we move towards the
other charge the net electric field becomes zero at x = a thereafter
the influence of charge at x = 2a dominates and net field increases
in negative direction of x-axis and grows unboundedly as we come
closer and closer to the charge at x = 2a.
5. (b) From figure, , where
or
or
or
6. (b) Let us consider a spherical shell of radius x and thickness dx.
Charge on this shell
dq = ?.4px
2
dx = 
? Total charge in the spherical region from centre to r (r < R) is
= 
= 
= 
? Electric field at r,  E = 
= 
= 
7. (a) The flux is zero according to Gauss’ Law because it is a open
surface which enclosed a charge q.
8. (b)
Electric field at C due to electric dipole
=  along OC
Electric field at C due to induced charge must be equal and opposite to
electric field due to dipole as net field at C is zero.
9. (a) ? = linear charge density;
Charge on elementary portion dx = ? dx.
Electric field at 
Horizontal electric field, i.e., perpendicular to AO, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of
AO
= 
Also, d? =   or  dx = ad?
E =   
=  = 
10. (d) Since ,
where q is the total charge.
As shown in the figure, flux associated with the curved surface B is f =
f
B
Let us assume flux linked with the plane surfaces A and C be
f
A
 = f
C
 = f’
Therefore,
Page 5


1. (a) Surface charge density (s) = 
So 
and 
2. (d) Since electric field  decreases inside water, therefore flux 
 also decreases.
3. (c) When a dipole is placed in a uniform electric field, two equal and
opposite forces act on it. Therefore, a torque acts which rotates the
dipole.
4. (a) For the distances close to the charge at x = 0 the field is very high
and is in positive direction of x-axis. As we move towards the
other charge the net electric field becomes zero at x = a thereafter
the influence of charge at x = 2a dominates and net field increases
in negative direction of x-axis and grows unboundedly as we come
closer and closer to the charge at x = 2a.
5. (b) From figure, , where
or
or
or
6. (b) Let us consider a spherical shell of radius x and thickness dx.
Charge on this shell
dq = ?.4px
2
dx = 
? Total charge in the spherical region from centre to r (r < R) is
= 
= 
= 
? Electric field at r,  E = 
= 
= 
7. (a) The flux is zero according to Gauss’ Law because it is a open
surface which enclosed a charge q.
8. (b)
Electric field at C due to electric dipole
=  along OC
Electric field at C due to induced charge must be equal and opposite to
electric field due to dipole as net field at C is zero.
9. (a) ? = linear charge density;
Charge on elementary portion dx = ? dx.
Electric field at 
Horizontal electric field, i.e., perpendicular to AO, will be cancelled.
Hence, net electric field = addition of all electrical fields in direction of
AO
= 
Also, d? =   or  dx = ad?
E =   
=  = 
10. (d) Since ,
where q is the total charge.
As shown in the figure, flux associated with the curved surface B is f =
f
B
Let us assume flux linked with the plane surfaces A and C be
f
A
 = f
C
 = f’
Therefore,
11. (d)
Apparently,   and .
12. (a) Since lines of force starts from A and ends at B, so A is +ve and B is
–ve. Lines of forces are more crowded near A, so A > B.
13. (c)
When the plates are charged, the net acceleration is,
 = 
? T 
?
14. (b) Force on charge q
1 
due to q
2
 is
Force on charge q
1
 due to q
3
 is
The X - component of the force (F
x
)  on