Question

Match the following:

Column-$\text{I}$	Column-$\text{II}$
$\left(\text{i}\right).$ Infinite sheet of charge having positive charges.	$\left(\text{a}\right).$ Uniform non-zero electric field intensity in the vicinity.
$\left(\text{ii}\right).$ At the center of Non-conducting uniformly charged solid sphere.	$\left(\text{b}\right).$ Uniform decreasing potential in the vicinity.
$\left(\text{iii}\right).$ Inside a positively charged conducting sphere taking infinity as a reference point.	$\left(\text{c}\right).$ maximum magnitude of potential.

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

Answer 1

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

For an infinite sheet of charge (Lets say in $x-y$ plane), Electric field is given by$$E=\frac{\sigma }{2{\epsilon }_0}$$
which does not depend upon the position of unit test charge, therefore electric field is uniform and nonzero as charge on the sheet is nonzero.

Since, $E=-\frac{dV}{dz}$

$\Rightarrow dV=-Edz$

Thus, Potential decreases uniformly along the direction of electric field.

For a uniformly charged non conducting solid sphere, Potential at any inside point is given by $$V_{in}=\frac{Q}{8\pi {\epsilon }_0{R}^3}(3R^2-r^2)$$At center $(r=0)$ $\therefore V_{in}=\frac{3Q}{8\pi {\epsilon }_{0}R}$

Which is the maximum value as $V_{center}=\frac{3}{2}\times Vsurface$

For a charged conducting sphere, assuming potential is zero at infinity.
Potential at surface of the sphere is given by $$V_{in}=\frac{Q}{4\pi {\epsilon }_{0}R}$$
Thus, we can say that, value of the potential is maximum at every point inside the conducting sphere.

Hence, option (a) is the correct answer.