Parallel Plate Capacitor: Practice Problem & Solution

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Parallel Plate Capacitor. Understanding the underlying formula is key to arriving at the correct answer below:

In a parallel plate capacitor made by stacking

$n$

equally spaced plates connected alternatively:

Key Points:

1. The plates connected alternatively form a series of capacitors.
2. If there are
   $n$ 

   plates, the number of gaps (capacitors in series) between them is $n - 1$ 

   .

For capacitors in series:

The total capacitance

$C_{\text{total}}$

is given by:

$$\frac{1}{C_{\text{total}}} = \frac{1}{C} + \frac{1}{C} + \dots + \frac{1}{C} \, \, (\text{n - 1 times})$$

$$\frac{1}{C_{\text{total}}} = \frac{n - 1}{C}$$

$$C_{\text{total}} = \frac{C}{n - 1}$$

However, because the plates are connected alternatively, these effectively act as

$n - 1$

capacitors in parallel.

For capacitors in parallel:

The equivalent capacitance is:

$$C_{\text{eq}} = (n - 1)C$$

Thus, the resultant capacitance is

$(n - 1)C$

.