Question

Consider a cube of uniform charge density $\rho$. The ratio of electrostatic potential at the centre of the cube to that at one of the corners of the cube is

• A. $2$
• B. $\sqrt{3}$/2
• C. $\sqrt{2}$
• D. $1$

Answer 1

Answer: (A)

For any charge distribution,
Potential $\alpha \frac{\text{charge}}{\text{Distance}}$
So, potential due to a cube at its corner is
$V=\frac{C.Q}{\alpha }$
where, $\alpha$ = side length, $Q$ = charge and $C$ = constant.
Now, consider point $A$ is centre of cube.
Potential at point $A=8\times$ Potential due to a cube of side a and charge density $\rho$.
$\therefore V_A=\frac{8C\cdot Q}{\alpha }=\frac{8C\cdot \rho \cdot {\alpha }^3}{\alpha }$
where, $\rho$ = charge density.
$\Rightarrow V_A=8C\rho \cdot {\alpha }^2$
Now, potential at point $B$ = potential due
to a cube of side $2_a$ and charge density $\rho$.
$\Rightarrow V_B=\frac{C\cdot Q}{2a}=\frac{C\cdot \rho \cdot {\left(2a\right)}^3}{2a}$
$\Rightarrow V_B=4C\cdot \rho \cdot {\alpha }^2$
So, required ratio is $\frac{V_A}{V_B}=\frac{8C\rho a^2}{4Cp{a}^2}=2$