First slide

Redistribution of charges and concept of common potential

Question

Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius $\frac{2}{3} R$ and $\frac{1}{3} R$ have $12 μ C$ and $- 3 μ C$ charges, respectively, and are at a large distance from each other. They are now connected by a conducting wire. A long time after this is done the charges on $S_{1}$ and $S_{2}$ are respectively :

Moderate

A

$6 μ C a n d 3 μ C$
B

$4.5 μ C o n b o t h$
C

$+ 4.5 μ C a n d - 4.5 μ C$
D

$3 μ C a n d 6 μ C$

Solution

After connected, Both spheres are at same potential.
$\begin{array}{l} \frac{q_{1}}{C_{1}} = \frac{q_{2}}{C_{2}} \\ Capacitance of sphere, C \propto R \\ T o t a l c h a r g e = Q = q_{1} + q_{2} \\ L e t s a y q_{1} = q, t h e n q_{2} = Q - q_{1} = (12 - 3) - q \\ \Rightarrow \frac{q}{2} = \frac{(12 - 3) - q}{1} \\ \Rightarrow q = 18 - 2 q \\ \Rightarrow q = 6 \\ H e n c e, \\ q_{1} = q = 6 μ C \\ q_{2} = Q - q_{1} = (12 - 3) - q = 9 - 6 = 3 μ C \end{array}$

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