Consider a cube of uniform charge density rho | vedclass

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Question

$(a)$ For any charge distribution,

Potential $\propto \frac{	ext { Charge }}{	ext { Distance }}$

So, potential due to a cube at its corner is

Vedclass · Accepted Answer

$2$

Vedclass · Answer

$(a)$ For any charge distribution,

Potential $\propto \frac{	ext { Charge }}{	ext { Distance }}$

So, potential due to a cube at its corner is

$V=\frac{C \cdot Q}{a}$

where, $a=$ side length, $Q=$ charge

and $C=$ constant.

Now, consider point $A$ is centre of cube.

Potential at point $A=8 	imes$ Potential due

to a cube of side $a$ and charge density $ho$.

$	herefore \quad V_A=\frac{8 C \cdot Q}{a}=\frac{8 C \cdot ho \cdot a^3}{a}$

where, $ho=$ charge density.

$\Rightarrow \quad V_A=8 C ho \cdot a^2 \quad \dots(i)$

Now, potential at point $B=$ potential due to a cube of side $2 a$ and charge density $ho$.

$\Rightarrow \quad V_B=\frac{C \cdot Q}{2 a}=\frac{C \cdot ho \cdot(2 a)^3}{2 a}$

$\Rightarrow \quad V_B=4 C \cdot ho \cdot a^2 \quad \dots(ii)$

So, required ratio is $\frac{V_A}{V_B}=\frac{8 C ho a^2}{4 C ho a^2}=2$.

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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