Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be

Question

Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be (A) 1:1 (B) 1:2 (C) 1:4 (D) 1:8

Tardigrade Admin · Accepted Answer

First, let us understand the argument of proportionality with respect to electric potential. The electric potential due to any continuous charge distribution at a field point is

V = \frac{1}{4 π ε _{0}} \int \frac{ρ ( d u )}{r}

where

ρ

is the volume charge density,

d u

is a small elemental volume and

r

is the distance of the field point from the charge distribution. So If

ρ

is constant, then the potential at a field point is directly proportional to the total volume or in other words the total charge of the continuous charge distribution. Similarly, the potential will be inversely proportional to the distance

r

.

Now, let us assume that the cube which is given in the problem (call it Full Cube) has an edge length

2 a

and we have made this cube by combining

8

smaller cubes (call them Half Cubes) of edge

a

.
If the potential at any of the corners of the Full Cube is

V_{cor n er} = V_{0}

then the potential at the corner of the Half Cube is

V_{cor n er}^{'} = V_{0} \times \frac{2}{8} = \frac{V _{0}}{4}

We have multiplied by a factor of two because all the distances in the Half Cube are half of the distances in the Full Cube and we have divided by eight because the charge of the Half Cube is one-eighth of the charge of the Full Cube.
Since the potential at a point is a scalar quantity, we can calculate the potential at the centre of the Full Cube by simply adding the potential created by the eight Half Cube.

V_{ce n t re} = 8 V_{cor n er}^{'}

\Rightarrow V_{ce n t re} = 8 \times \frac{V _{0}}{4} = 2 V_{0}

Therefore the required ratio for the Full Cube

\frac{V _{cor n er}}{V _{ce n t re}} = \frac{V _{0}}{2 V _{0}} = \frac{1}{2}

Q. Consider a solid cube made up of insulating material having a uniform volume charge density. Assuming the electrostatic potential to be zero at infinity, the ratio of the potential at a corner of the cube to that at the centre will be

$1 : 1$

$1 : 2$

$1 : 4$

$1 : 8$