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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» 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)`. |
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