Two concentric shells of radii `R` and `2 R` are shown in (Fig. 3.115). Initially, a charge q is imparted to the inner shells. Now, key `K_1` is closed and opened and then key `K_2` is closed and opened. After the keys `K_1 and K_2` are alterbately closed n times each, find the potential difference between the shells. Note that finally key `K_2` remains closed.
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Answer

`(-q)/(4 pi epsilon_0 2^(n + 1) R)`

Explanation

When `K_1` is closed first time, the outer sphere is earthed and the potential on it becomes zero. Let the charge on it be `q_1`.
Potential due to charge on the inner sphere and that due to charge on the outer sphere is
`V'_1 = (1)/(4 pi epsilon_0)[q/(2 R) + (q'_1)/(2 R)] = 0` or `q'_1 = -q`
When `K_2` is closed first time, the potential `V'_2` on the inner sphere becomes zero as it earthed. Let the new charge on the inner sphere be`q'_2`.
`0 = (1)/(4 pi epsilon_0) (q'_2)/(R)+ (1)/(4 pi epsilon_0) ((- q))/((2 R))` or `q'_2 = q/(2)`
Now, when `K_1` will be closed second time, charge on the outer sphere will be `-q'_2`, i.e., `- q//2`. After one event involving closure and opening of `K_1` and `K_2`, charge is reduced to half of its initial value. Similarly, when `K_1` will be closed `n^(th)` time, charge on the outer sphere will be `-q //(2^(n - 1))` as each time charge will be reduced to half of the previous value.
After closing `K_2 n^(th)` time, charge on the inner shell will be negative of half the charge on the outer shell, i.e., `(+ q//2^n)` and potential on it will be zero. For potential of the outer shell
`V_0 = (1)/(4 pi epsilon_0) ((+ q //2^n))/(2 R) + (1)/(4 pi epsilon_0) ((-q //2^(n - 1)))/(2 R)`
=`(- q[- 1 + 2])/(4 pi epsilon_0 2^(n +1) R)= (-q)/(4 pi epsilon_0 2^(n -1) R)`
Potential difference is
`V_0 - V_i = (-q)/(4 pi epsilon_0 2^(n+1) R) - 0 = (-q)/(4 pi epsilon_0 2^(n + 1) R)`.