Question

Consider a solid cube of uniform charge density of insulating material. What is the ratio of the electrostatic potential at a corner to that at the centre. (Take the potential to be zero at infinity, as usual )

• A. $\frac{1}{1}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{9}$

Answer 1

Answer: (B)

$\frac{1}{2}$

Answer 2

r- charge density of the cube

V_l^corner = potential at the corner of a cube of side l.

V_l^centre = potential at the centre of a cube of side l.

V_l/2^centre = potential at the centre of a cube of side $\frac{\mathcal{l}}{2}$.

V_l/2^corner = potential at the corner of a cube of side $\frac{\mathcal{l}}{2}$.

By dimensional analysis V_l^corner µ $\frac{Q}{\mathcal{l}} = \rho\mathcal{l}^{2}$

V_l^corner = 4 V_l/2^corner

But by super position V_l^centre = 8 V_l/2^corner because the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made

Therefore $\frac{V_{\mathcal{l}}^{corner}}{V_{\mathcal{l}}^{centre}} = \frac{4V_{\mathcal{l}/2}^{coner}}{8V_{\mathcal{l}/2}^{centre}} = \frac{1}{2}$