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Partially Filled Dielectrics

The figure sh...

Question

The figure shows two capacitors in series, the rigid central conducting section of length $b$ being movable vertically. If the potential of the upper and lower plates are $V_{1}$ and $V_{2}$ , respectively, then the potential of the rigid section is :

V_{1} - \frac{(V_{1} - V_{2}) x}{(a - b)}

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V_{1} - \frac{(V_{2} - V_{1}) x}{(a - b)}

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V_{1} - \frac{(V_{1} - V_{2}) x}{(a + b)}

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V_{1} - \frac{(V_{2} - V_{1}) x}{(a + b)}

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Solution

The correct option is A $V_{1} - \frac{(V_{1} - V_{2}) x}{(a - b)}$
The circuit is equivalent to two capacitors in series.
$C_{1} = \frac{ε_{0} A}{x}$
$C_{2} = \frac{ε_{0} A}{a - b - x}$
Now potential will be distributed in these capacitors.
We can reduce this structure in the following circuit.
$V_{1} - V = (V_{1} - V_{2}) \frac{C_{2}}{C_{1} + C_{2}}$
$V_{1} - V = (V_{1} - V_{2}) \frac{\frac{ε_{0} A}{a - b - x}}{\frac{ε_{0} A}{x} + \frac{ε_{0} A}{a - b - x}}$ $= (V_{1} - V_{2}) \frac{x}{(a - b)}$
$V = V_{1} - (V_{1} - V_{2}) \frac{x}{(a - b)}$

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The figure shows two capacitors in series, the rigid central conducting section of length b being movable vertically. If the potential of the upper and lower plates are V1 and V2, respectively, then the potential of the rigid section is :

The figure shows two capacitors in series, the rigid central conducting section of length $b$ being movable vertically. If the potential of the upper and lower plates are $V_{1}$ and $V_{2}$ , respectively, then the potential of the rigid section is :