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2. Electric Potential and Capacitance
normal
Potential difference between centre $\&$ the surface of sphere of radius $R$ and uniform volume charge density $\rho$ within it will be :
A
$\frac{{\rho \,{R^2}}}{{6\,{ \in _0}}}$
B
$\frac{{\rho \,{R^2}}}{{4\,{ \in _0}}}$
C
$0$
D
$\frac{{\rho \,{R^2}}}{{2\,{ \in _0}}}$
Solution
$\rho=\frac{a}{(4 / 3) \pi R^{3}}$
$\therefore q=\frac{4}{3} \pi \rho R^{3}$
$V_{C}+V_{S}=\frac{3}{2}\left(\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}\right)-\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{R}$
$=\frac{q}{8 \pi \varepsilon_{0} R}$
Substituting the value of $q$ we have
$V_{C}-V_{S}=\frac{\rho R^{3}}{6 \varepsilon_{0}}$
Standard 12
Physics
