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Question

$ho=\frac{a}{(4 / 3) \pi R^{3}}$

$	herefore q=\frac{4}{3} \pi ho R^{3}$

$V_{C}+V_{S}=\frac{3}{2}\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot

Vedclass · Accepted Answer

$\frac{{\rho \,{R^2}}}{{6\,{ \in _0}}}$

Vedclass · Answer

$ho=\frac{a}{(4 / 3) \pi R^{3}}$

$	herefore q=\frac{4}{3} \pi ho R^{3}$

$V_{C}+V_{S}=\frac{3}{2}\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}ight)-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$

$=\frac{q}{8 \pi \varepsilon_{0} R}$

Substituting the value of $q$ we have

$V_{C}-V_{S}=\frac{ho R^{3}}{6 \varepsilon_{0}}$

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

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