Question

Match the following:

Column-$\text{I}$	Column-$\text{II}$
$\left(\text{i}\right).$ Positively charged Spherical conductor.	$\left(\text{a}\right).$ At the surface , electric field is continuous and maximum.
$\left(\text{ii}\right).$ Spherical non conductor having uniform volume charge distribution.	$\left(\text{b}\right).$ at the surface, electric field is dis-continuous.
$\left(\text{iii}\right).$ Positively charged ring.	$\left(\text{c}\right).$ Electric field is uniform.
$\left(\text{iv}\right).$ Infinite thin sheet of positive charge	$\left(\text{d}\right).$ at the center, electric field is zero.

• A. $\text{i}\to \text{a,b},\text{ii}\to \text{d},\text{iii}\to \text{a},\text{iv}\to \text{c}$
• B. $\text{i}\to \text{a,c},\text{ii}\to \text{a,d},\text{iii}\to \text{d},\text{iv}\to \text{c}$
• C. $\text{i}\to \text{b,d},\text{ii}\to \text{a},\text{iii}\to \text{c},\text{iv}\to \text{d}$
• D. $\text{i}\to \text{b,d},\text{ii}\to \text{a,d},\text{iii}\to \text{d},\text{iv}\to \text{c}$

Answer 1

The correct option is D $\text{i}\to \text{b,d},\text{ii}\to \text{a,d},\text{iii}\to \text{d},\text{iv}\to \text{c}$

In the case of hollow sphere or Solid conducting sphere of radius $R$., electric field is zero at any internal point $r<R$.

At the surface, $E$ is either zero (minimum) or $\frac{Q}{4\pi {\epsilon }_0{R}^2}$ (maximum).

So we can say that , $\text{(i)}\to \text{b,d}$

In the case of spherical volume distribution of charge,

Electric field inside the sphere is given by $$E_{in}=\frac{Qr}{4\pi {\epsilon }_0{R}^3}$$
From this, it is clear that, electric field is zero at the center and at the surface $(r=R)$ , $E$ is continuous and maximum $(=\frac{Q}{4\pi {\epsilon }_0{R}^2})$.

So we can say that, $\text{(ii)}\to \text{a,d}$

For a charged ring, electric field at any point $x$ along the axis of the ring is given by $$E_{axis}=\frac{Qx}{4\pi {\epsilon }_0{(x^2+a^2)}^{3/2}}$$
At center, $x=0$ $$\therefore E_{center}=0$$
So we can say that, $\text{(iii)}\to \text{d}$

For infinite thin sheet of charge $Q$, electric field at any point is given by $$E=\frac{\sigma }{2{\epsilon }_0}$$
Which doesnt depend upon the position of unit test charge, therefore electric field is uniform.

So we can say that, $\text{(iv)}\to \text{c}$

Hence, option (d) is the correct answer.