Consider a cube of uniform charge density rho | vedclass

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(A) Let the side length of the small cube be $a$. The large cube has a side length of $2a$ and is composed of $8$ such small cubes.Let $V_c$ be the po

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$2$

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(A) Let the side length of the small cube be $a$. The large cube has a side length of $2a$ and is composed of $8$ such small cubes.Let $V_c$ be the potential at the centre of a small cube of side $a$ and charge density $ho$. The total potential at the centre of the large cube (point $A$) is the sum of potentials due to the $8$ small cubes surrounding it.Since the centre of the large cube is the common corner of the $8$ small cubes,the potential at the centre $A$ is $V_A = 8 	imes V_{	ext{corner of small cube}}$.For a small cube of side $a$,the potential at its corner due to its own charge $Q = ho a^3$ is $V_{	ext{corner}} = k \frac{Q}{a} = k ho a^2$ (where $k$ is a constant).Thus,$V_A = 8 	imes (k ho a^2) = 8 k ho a^2$.Now,consider the potential at the corner $B$ of the large cube of side $2a$. The potential at the corner of a cube of side $L$ and charge density $ho$ is proportional to $ho L^2$.For the large cube of side $2a$,$V_B = k ho (2a)^2 = 4 k ho a^2$.The ratio of the potential at the centre to that at the corner is $\frac{V_A}{V_B} = \frac{8 k ho a^2}{4 k ho a^2} = 2$.

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

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