ExamGOAL
Books
PreviousNext
11
MCQ (Multiple Correct Answer)

Consider a region inside which there are various types of charges but the total charge is zero. At points outside the region,

A
the electric field is necessarily zero
B
the electric field is due to the dipole moment of the charge distribution only
C
the dominant electric field is $\propto \frac{1}{r_3}$, for large $r$, where $r$ is the distance from a origin in this regions
D
the work done to move a charged particle along a closed path, away from the region, will be zero
12
MCQ (Multiple Correct Answer)

Refer to the arrangement of charges in figure and a Gaussian surface of radius $R$ with $Q$ at the centre. Then,

A
total flux through the surface of the sphere is $\frac{-Q}{\varepsilon_0}$
B
field on the surface of the sphere is $\frac{-Q}{4 \pi \varepsilon_0 R^2}$
C
flux through the surface of sphere due to $5 Q$ is zero
D
field on the surface of sphere due to $-2 Q$ is same everywhere
13
MCQ (Multiple Correct Answer)

A positive charge $Q$ is uniformly distributed along a circular ring of radius R.A small test charge $q$ is placed at the centre of the ring figure. Then,

A
if $q>0$ and is displaced away from the centre in the plane of the ring, it will be pushed back towards the centre
B
if $q<0$ and is displaced away from the centre in the plane of the ring, it will never return to the centre and will continue moving till it hits the ring
C
if $q<0$, it will perform SHM for small displacement along the axis
D
$q$ at the centre of the ring is in an unstable equilibrium within the plane of the ring for $q>0$
14
Subjective

An arbitrary surface encloses a dipole. What is the electric flux through this surface?

Explanation

From Gauss' law, the electric flux through an enclosed surface is given by $\oint_s E . d S=\frac{q}{\varepsilon_0}$.

Here, $q$ is the net charge inside that enclosed surface.

Now, the net charge on a dipole is given by $-q+q=0$

$\therefore \quad$ Electric flux through a surface enclosing a dipole $=\frac{-q+q}{\varepsilon_0}=\frac{0}{\varepsilon_0}=0$

15
Subjective

A metallic spherical shell has an inner radius $R_1$ and outer radius $R_2$. A charge $Q$ is placed at the centre of the spherical cavity. What will be surface charge density on

(i) the inner surface

(ii) the outer surface?

Explanation

Here, the charge placed at the centre of the spherical cavity is positively charged. So, the charge created at the inner surface of the sphere, due to induction will be $-Q$ and due to this charge created at outer surface of the sphere is $+Q$.

Now, surface charge density on the inner surface $=\frac{-Q}{4 \pi R_1^2}$

and Surface charge density on the inner surface $=\frac{+Q}{4 \pi R_2^2}$

PreviousNext