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(B) Let the three plates be $1, 2,$ and $3$ from left to right. Plate $1$ has charge $+q$ and plate $3$ has charge $-2q$. Plate $2$ is initially neutr

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$-q$

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(B) Let the three plates be $1, 2,$ and $3$ from left to right. Plate $1$ has charge $+q$ and plate $3$ has charge $-2q$. Plate $2$ is initially neutral (charge $0$).When switch $S$ is closed, plate $2$ is connected to the earth, so its potential becomes $0$.Let the charge on the inner surfaces of plate $2$ be $q_2'$ and $q_2''$ facing plate $1$ and plate $3$ respectively.The potential of plate $2$ is given by the sum of potentials due to all charges: $V_2 = \frac{1}{4\pi\epsilon_0 A} [q_1 d_1 + q_2 d_2 + q_3 d_3] = 0$, where $d$ is the distance.More simply, for a grounded plate between two charged plates, the induced charge on the grounded plate is such that the potential at its location is zero.The electric field due to plate $1$ $(+q)$ at the position of plate $2$ is $E_1 = \frac{q}{2A\epsilon_0}$ (directed to the right).The electric field due to plate $3$ $(-2q)$ at the position of plate $2$ is $E_3 = \frac{2q}{2A\epsilon_0}$ (directed to the right).The total electric field at plate $2$ due to plates $1$ and $3$ is $E_{net} = \frac{q}{2A\epsilon_0} + \frac{2q}{2A\epsilon_0} = \frac{3q}{2A\epsilon_0}$ (to the right).To make the potential zero, the charge $Q_{induced}$ on the grounded plate must create an opposing field. The charge on the plate will be $Q_{induced} = -(	ext{Total charge induced by external fields}) = -(\frac{q}{2} + \frac{-2q}{2}) = -(-\frac{q}{2}) = +\frac{q}{2}$? No, using the property of parallel plates: the charge on the inner faces of the grounded plate will be $-q/2$ and $+2q/2 = +q$. The total charge on the plate is $Q = -q/2 + q = +q/2$. Wait, the charge flowing to the earth is the charge that was on the plate minus the final charge. Since the plate was neutral $(0)$ and the final charge is $+q/2$, the charge that flows to the earth is $-q/2$. Let's re-evaluate: The potential at the middle plate due to plate $1$ is $-E_1 d_1$ and due to plate $3$ is $-E_3 d_3$. For $V=0$, the induced charge $Q$ must satisfy $Q/C_{eq} + V_{ext} = 0$. The charge on the middle plate will be $Q = - \frac{C_1 V_1 + C_3 V_3}{C_1 + C_3}$. Given the symmetry and standard results for this configuration, the charge on the middle plate becomes $+q/2$. Since it was $0$, the charge that flows from the earth to the plate is $+q/2$, meaning charge $-q/2$ flows to the earth.

The metal plate on the left in the figure has a charge of $+q$. The metal plate on the right has a charge of $-2q$. What charge will flow to the earth when switch $S$ is closed, given that the middle plate is initially neutral?

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