Two concentric spherical shells of radius R_1 | vedclass

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(D) Let the charge on the inner shell after closing the key be $q_1'$.Since the inner shell is connected to the ground, its potential becomes zero.The

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$-\left( q_1 + q_2 \frac{R_1}{R_2} \right)$

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(D) Let the charge on the inner shell after closing the key be $q_1'$.Since the inner shell is connected to the ground, its potential becomes zero.The potential of the inner shell is given by the sum of potentials due to its own charge and the charge on the outer shell:$V_1 = \frac{k q_1'}{R_1} + \frac{k q_2}{R_2} = 0$From this, we get:$\frac{q_1'}{R_1} = -\frac{q_2}{R_2} \implies q_1' = -q_2 \frac{R_1}{R_2}$The charge that flows through the key is the change in charge on the inner shell:$\Delta q = q_{final} - q_{initial} = q_1' - q_1$$\Delta q = -q_2 \frac{R_1}{R_2} - q_1 = -(q_1 + q_2 \frac{R_1}{R_2})$Thus, the magnitude of charge flowing to the ground is $(q_1 + q_2 \frac{R_1}{R_2})$.

Two concentric spherical shells of radius $R_1$ and $R_2$ have $q_1$ and $q_2$ charge respectively as shown in the figure. How much charge will flow through key $k$ when it is closed?

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