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Exactly midway of a parallel plate capacitor a charged plate was placed having charge q. The capacitance is connected to a battery of voltage V. The plate is moved perpendicular to itself by a distance x, What is the charge flowing through the external circuit of the capacitor, if the distance between its plates equal to d?

Exactly midway of a parallel plate capacitor a charged plate was placed having charge q. The capacitance is connected to a battery of voltage V. The plate is moved perpendicular to itself by a distance x, What is the charge flowing through the external circuit of the capacitor, if the distance between its plates equal to d?

Grade:11

1 Answers

Eshan
askIITians Faculty 2095 Points
2 years ago
The plate kept between a capacitor divides the capacitor into two capacitors connected in series.

For the intial configuration, the capacitance of each half of the capacitor is\dfrac{A\epsilon_0}{d/2}=\dfrac{2A\epsilon_0}{d}
Hence when connected the two capacitors are connected in series, the overall capacitance becomes\dfrac{A\epsilon_0}{d}.

When the plate is moved by distancex, the capacitane of one side of the capacitor becomesC_1=\dfrac{A\epsilon_0}{\dfrac{d}{2}-x}
The capacitance of other half becomesC_2=\dfrac{A\epsilon_0}{\dfrac{d}{2}+x}
Hence the total capacitance becomes\dfrac{C_1C_2}{C_1+C_2}=\dfrac{A\epsilon_0}{d}
Therefore, the capacitance of the system does not change even by moving the plate. So, the charge over the capacitor does not change at any time and so no current flows through the external circuit.

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