Question

A parallel-plate capacitor is connected to a battery as shown in the figure. A conducting plate is inserted mid-way between the two plates. Take the plate area as $A$. The distance between each plate in the new configuration is $d$.

If the middle and upper plates are short circuited, the extra charge flown through the battery is

• A. $\frac{{\epsilon }_{0}AV}{d}$
• B. Zero
• C. $\frac{{\epsilon }_{0}AV}{2d}$
• D. $\frac{2{\epsilon }_{0}AV}{d}$

Answer 1

The correct option is C $\frac{{\epsilon }_{0}AV}{2d}$

We have,
$C_i=\frac{{\epsilon }_{0}A}{2d}$

$Q_i=C_i{V}_i$
$Q_i=\frac{{\epsilon }_{0}A}{2d}\times V$

Now upper plates are short circuited,

So, only one plate remains in picture and the potential difference across it is $V$.
$C_f=\frac{{\epsilon }_{0}A}{d}$
$Q_f=C_f{V}_f$
$Q_f=\frac{{\epsilon }_{0}A}{d}\times V$

So, the extra charge flown is
$Q_f-Q_i$
$\frac{{\epsilon }_{0}A}{d}V-\frac{{\epsilon }_{0}AV}{2d}$
$=\frac{{\epsilon }_{0}AV}{2d}$