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Q.
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
KVPYKVPY 2016
Solution:
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 $\frac{V_{A}}{V_{B}}=\frac{8C\rho a^{2}}{4Cpa^{2}}=2$
