In the system shown middle plate is having a charge $+ Q$ , the other plates are charge neutral initially and both the switches are open. If we close both the switches simultaneously find out the final charge on the inner side of the left most plate.

\frac{- Q d_{1}}{d_{1} + d_{2}}

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\frac{- Q}{2}

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\frac{- Q d_{2}}{d_{1} + d_{2}}

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zero

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Solution

The correct option is C $\frac{- Q d_{2}}{d_{1} + d_{2}}$
When both switch is closed, charge on outer surface of both the plate will become zero. Let the final charge on left plate be $Q_{1}$ and on right plate be $Q_{2}$ .

$- (Q_{1} + Q_{2}) = Q$
$\Rightarrow Q_{1} + Q_{2} = - Q$ ....(1)

Since both the plates are grounded, potential difference between the plate $A$ and $C$ will be zero.

$V_{B A} = \frac{Q_{1}}{A ϵ_{o}} d_{1}$ and $V_{B C} = \frac{Q_{2}}{A ϵ_{o}} d_{2}$

$∵ V_{B A} + V_{C B} = 0$
$\Rightarrow V_{B A} - V_{B C} = 0$

$\Rightarrow \frac{Q_{1}}{A ϵ_{o}} d_{1} - \frac{Q_{2}}{A ϵ_{o}} d_{2} = 0$

$\Rightarrow Q_{2} = \frac{Q_{1} d_{1}}{d_{2}}$

Now, using equation (1):
$Q_{1} + \frac{Q_{1} d_{1}}{d_{2}} = - Q$

$\Rightarrow Q_{1} = \frac{- Q d_{2}}{d_{1} + d_{2}}$

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In the system shown middle plate is having a charge +Q, the other plates are charge neutral initially and both the switches are open. If we close both the switches simultaneously find out the final charge on the inner side of the left most plate.

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