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
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