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A parallel plate capacitor is charged as shown (Q is given). A metal slab with the total charge +Q is placed inside the capacitor as shown. The thickness of the slab is d. The distance between the top plate and the top of the slab is 2d, and the distance between the bottom plate and the botthom of the slab is d. Each plate is grounded through a galvanometer as shown . Find the charge that passes through each galvanometer after both switches are closed simultaneously. |
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Answer» A total charge of `-Q//3` flows `(through G_1)` from the bottom PLATE to ground. Initial charges on each surface  Chare on OUTER surfaces `q_(a)=q_(b)=(Q+Q-Q)/(2)=(Q)/(2)` and `q_(b)=(Q-(Q-Q))/(2)=(Q)/(2)` `C_(1)=C=(epsilon_(0)A)/(2d)` or `C_(2)=2C=(epsilon_(0)A)/(d)`  When both switches are CLOSED the circuit can be redrawn as SHOWN. Moving from `A` and `B` `Q+(x)/(C)-((Q0x))/(2C)=O` or `x=(Q)/(3)`  Final charges on each plate will be Hence charge moving from top plate to ground is `Q-(-(Q)/(3))` `=(4Q)/(3)` and from bottom plate to gound is `-Q-(-(2Q)/(3))=(Q)/(3)` |
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