Question

2. The conducting shells A and B are arranged as
shown below. If charge on the shell B is q then
electric flux linked with the spherical gaussian
surface S is
​

Answer 1

Answer:

Explanation: this is the solution

Answer 2

Answer: Flux through surface S is -q/2∈₀

Explanation:

Let Δq charge flows from ground to shell A.

Shell A is connected to ground hence its potential should be 0.

Potential of shell A = Potential due to charge on A +  Potential due to charge on B

Potential of shell A = k Δq/r + kq/2r

Therefore,

⇒  k Δq/r + kq/2r = 0

⇒  k Δq/r + kq/2r = 0

⇒ 2Δq + q = 0

⇒ Δq = -q/2

Flux through surface S = (Charge enclosed by surface S)/∈₀

Flux through surface S = Δq/∈₀

Flux through surface S = -q/2∈₀

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