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Q.
A charge $Q$ is distributed over two concentric hollow spheres of radii $a$ and $b(a>b)$, so that the surface charge densities are equal. The potential at the common centre is $\frac{1}{4 \pi \varepsilon_{0}}$ times
KEAMKEAM 2015Electrostatic Potential and Capacitance
Solution:
As, charge $O$ is distributed over two concentric hollow spheres of radii $a$ and $b(a>b)$,
$Q=4 \pi a^{2} \sigma+4 \pi b^{2} \sigma$
$\sigma=\frac{Q}{4 \pi\left(a^{2}+b^{2}\right)}$
and potential energy, we have
$U=\frac{K Q_{1}}{a}+\frac{K Q_{2}}{b}$
As, potential in common centre,
$K=\left(\frac{4 \pi a^{2} \sigma}{a}+\frac{4 \pi b^{2} \sigma}{b}\right)$
$=K(4 \pi a \sigma+4 \pi b \sigma)=K \sigma(a+ b)$
$=K \cdot 4 \pi\left(\frac{Q}{4 \pi\left(a^{2}+b^{2}\right)}\right) \cdot(a +b)$
$=\frac{K \cdot Q(a +b)}{a^{2}+b^{2}}=K Q\left(\frac{a +b}{a^{2}+b^{2}}\right)$
Here, $K=\frac{1}{4 \pi \varepsilon_{0}}$