Two small conducting balls A and B of radius r1 and r2 have charges q1 and q2 respectively. They are connected by a wire.Obtain the expression for charges on A and B, in equilibrium
Question
Two small conducting balls A and B of radius r1 and r2 have charges q1 and q2 respectively. They are connected by a wire.Obtain the expression for charges on A and B, in equilibrium
Solution
When two conducting balls A and B are connected by a wire, the charges on them will redistribute until they reach an equilibrium state. In this state, the potential on both spheres will be the same.
The potential V of a sphere with charge Q and radius R is given by the formula V = Q/(4πεR), where ε is the permittivity of free space.
For sphere A, we have:
V1 = q1/(4πεr1) ----(1)
And for sphere B:
V2 = q2/(4πεr2) ----(2)
At equilibrium, the potentials of the two spheres are equal, so we can set V1 = V2:
q1/(4πεr1) = q2/(4πεr2)
Solving this equation for q1 gives:
q1 = q2 * r1/r2 ----(3)
Similarly, solving for q2 gives:
q2 = q1 * r2/r1 ----(4)
These are the expressions for the charges on spheres A and B in equilibrium. Note that the total charge Q = q1 + q2 is conserved during this process.
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