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Q. The figure shows four identical conducting plates each of area $A$. The separation between the consecutive plates is equal to $L$. When both the switches are closed, the charge present on the upper surface of the lowest plate from the top is written as $\frac{x V_{0} \varepsilon_{0} A}{L}$. What is the value of $x ?$
(Treat symbols as having usual meaning.)Physics Question Image

Electrostatic Potential and Capacitance

Solution:

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$C\left(x-V_{0}\right)+C(x-0)+C\left(x+4 V_{0}\right)=0$
$3 C x=-3 C V_{0}$
$Q=\frac{3 V_{0} \varepsilon_{0} A}{I}$
$\Rightarrow x=3$