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Q.

Potential difference between centre & the surface of sphere of radius R and uniform volume charge density $ρ$ within it will be:

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a

0

b

$\frac{{ρR}^{2}}{6 ϵ_{0}}$

c

$\frac{{ρR}^{2}}{4 ϵ_{0}}$

d

$\frac{{ρR}^{2}}{2 ϵ_{0}}$

answer is A.

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Detailed Solution

Potential at surface is:

Question Image

$V_{s} = \frac{ρ \times 4 {πR}^{3}}{3 \times 4 {πϵ}_{0} R} = \frac{{ρR}^{2}}{3 ϵ_{0}}$

Potential at center = $\int dV = \int_{0}^{R} \frac{ρ 4 {πx}^{2} dx}{4 {πϵ}_{0} x}$

$\Rightarrow V_{C} = \frac{{ρR}^{2}}{2 ϵ_{0}} \Rightarrow V_{C} - V_{S} = \frac{{ρR}^{2}}{6 ϵ_{0}}$

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Potential difference between centre & the surface of sphere of radius R and uniform volume charge density ρ within it will be: