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Two conducting spheres having radii a and b are charged to $q_1$ and $q_2$ respectively. Find the potential difference between spheres.

1 Answer

$v_1= \large\frac{1}{4\pi \in_0} \frac{q_1}{a} +\frac{1}{4 \pi \in_0}\frac{q_2}{b} $ ......... (a)
The potential on the surface of the sphere 2 is given by,
$v=v_1 - v_2$
$V_2=\large\frac{1}{4 \pi \in_0} \frac{q_1}{b} +\frac{1}{4 \pi \in_0} \frac{q_2}{b} $
=> $ v= \large\frac{1}{4\pi \in_0} \frac{q_1}{a} -\frac{1}{4 \pi \in_0 } \frac{q_1}{b}$
=>$ v= \large\frac{q_1}{4 \pi \in_0} \bigg( \large\frac{1}{a}-\frac{1}{b} \bigg)$
answered Jun 7, 2014 by meena.p

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