A metal sphere A of radius a is charged to a potential V. It is enclosed inside a hollow metal shell B of radius b.

What will be the potential of A if it is connected to shell B by a conducting wire?

Show the working.

To find the potential of sphere A after it is connected to shell B by a conducting wire, we'll follow these steps:

Understanding Potential of Conducting Surfaces: When two conducting surfaces are connected by a wire, they will come to the same potential. This is due to the fact that charges will redistribute until both surfaces have the same electrical potential. This means that the potential will be equal on the sphere A and the shell B once they are connected.

Initial Potentials:

Sphere A, initially has a potential V and a radius a .
The shell B is initially uncharged and has a radius b . Its initial potential can be considered as 0 if not specified otherwise.


Using Charge Conservation: When a conducting wire connects both, the total charge is conserved. If Q A ​ is the initial charge on sphere A and Q B ​ is the initial charge on shell B. The final charge distribution ensures that:
Q A ​ + Q B ​ = Q A ′ ​ + Q B ′ ​
where Q A ′ ​ and Q B ′ ​ are the final charges on A and B respectively.

Calculating Final Potential: Once connected, the potential on both A and B becomes the same, say V f ​ . The potential on a conducting sphere is given by the formula:
V = r k Q ​
Hence, the potential for sphere A becomes:
V f ​ = a k Q A ′ ​ ​
and for shell B:
V f ​ = b k Q B ′ ​ ​
Since V f ​ is same for both A and B after they are connected:
a k Q A ′ ​ ​ = b k Q B ′ ​ ​

Solving for Charges and Potential: Using the fact that the total charge is conserved and that we have two equations (one from charge conservation and one from equal potentials), it is possible to solve for Q A ′ ​ and Q B ′ ​ and then calculate V f ​ .

Final Result: The mathematical problem is typically solved to find that the final potential V f ​ is:
V f ​ = 4 π ε 0 ​ Q t o t a l ​ ​ b 1 ​
Here, Q t o t a l ​ is the total charge system (i.e., the initial charge on sphere A assuming shell B was uncharged). This potential is uniform over both sphere A and shell B because they are conductors sharing charge through a conducting wire.


Overall, the ability of the electrons to move freely between the sphere and the shell results in the potential being uniformly equal.

Answered by danjohnbrain | 2025-07-06