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. $1:8$

Answer 1

Answer: (B)

$\rho -$ Charge density of the cube
$V_{l}^{c o r n e r}-$ Potential at the corner of a cube of side $l$
$V_{l}^{c e n t r e}-$ Potential at the centre of a cube of side $l$
$V_{l / 2}^{c e n t r e}-$ Potential at the center of a cube of side $\frac{l}{2}$
$V_{l / 2}^{c o r n e r}-$ Potential at the corner of a cube of side $\frac{l}{2}$
By dimensional analysis
$V_{l}^{c o r n e r} \propto \frac{Q}{l}=\rho l^{2}$
$V_{l}^{c o r n e r}=4 \, V_{l / 2}^{c o r n e r}$
But by superposition $V_{l}^{c e n t r e}=8V_{l / 2}^{c o r n e r}$
Because of the centre of the larger cube lies at a corner of the eight smaller cubes of which it is made
Therefore, $\frac{V_{l}^{c o r n e r}}{V_{l}^{c e n t r e}}=\frac{4 V_{l / 2}^{c o r n e r}}{8 V_{l / 2}^{c o r n e r}}=\frac{1}{2}$