Question

Obtain the expression for capacity of a parallel plate capacitor. How the capacity will be increased?

Hint:

Remember the three factors affecting capacitance: Area, Distance, and Medium. An easy way to recall this is the formula \(C = \frac{KA\epsilon_0}{d}\). To increase C, you increase the numerator factors (K, A) or decrease the denominator factor (d).

Answer 1

Step 1: Understanding the Concept:
A parallel plate capacitor consists of two large, flat, parallel conducting plates, each with an area A, separated by a small distance d. Capacitance (C) is a measure of its ability to store electric charge, defined as the ratio of the charge (Q) on one plate to the potential difference (V) between the plates.

Step 2: Derivation of the Expression:
Let the two plates of the capacitor have a surface charge density of \(+\sigma\) and \(-\sigma\). The charge on each plate is \(Q = \sigma A\).
Assuming the distance d is small compared to the dimensions of the plates, the electric field (\(E\)) between them is uniform and is given by: \[ E = \frac{\sigma}{\epsilon_0} = \frac{Q}{A\epsilon_0} \] where \(\epsilon_0\) is the permittivity of free space.
The potential difference (\(V\)) between the plates is the work done in moving a unit charge from one plate to another, which is \(V = E \times d\). \[ V = \left(\frac{Q}{A\epsilon_0}\right)d = \frac{Qd}{A\epsilon_0} \] The capacitance C is defined as \(C = \frac{Q}{V}\). Substituting the expression for V: \[ C = \frac{Q}{\frac{Qd}{A\epsilon_0}} = \frac{A\epsilon_0}{d} \] This is the expression for the capacitance of a parallel plate capacitor with a vacuum (or air) between the plates.
If a dielectric material with a dielectric constant K is inserted between the plates, the capacitance becomes: \[ C = \frac{KA\epsilon_0}{d} \]

Step 3: How to Increase the Capacity:
Based on the formula \(C = \frac{KA\epsilon_0}{d}\), the capacitance can be increased in the following ways: \begin{enumerate} \item By increasing the area of the plates (A): Capacitance is directly proportional to the area of the plates. \item By decreasing the distance between the plates (d): Capacitance is inversely proportional to the separation between the plates. \item By introducing a dielectric medium: Inserting a dielectric material with a high dielectric constant (K > 1) between the plates increases the capacitance by a factor of K. \end{enumerate}