Question

Define the capacity of a conductor. Derive an expression for the capacity of a parallel plate conductor. How can its capacity be increased?

Answer 1

Capacity is proportional to the charge stored and inversely proportional to the voltage across the plates. For a parallel plate capacitor, two plates, parallel to each other have opposite charges and are separated by a distance.

Formula Used: The formulae used in the solution are given here.
Capacity of a condenser $C = \dfrac{Q}{V}$ where $Q$ is the charge on the condenser and $V$ is the potential difference across its plates.

Complete Step by Step Solution
The capacitor is a component which has the ability or capacity to store energy in the form of an electrical charge producing a potential difference (Static Voltage) across its plates, much like a small rechargeable battery.
The charges required to produce a certain difference of potential between the plates of a condenser is a constant ratio to the potential. This constant ratio is called the capacity of condenser.
Mathematically, the capacity of a condenser $C = \dfrac{Q}{V}$ where $Q$ is the charge on the condenser and $V$ is the potential difference across its plates.
Consider a parallel plate capacitor having two plane metallic plates M and N placed parallel to each other. The plates carry equal and opposite charges $+ Q$ and $- Q$ respectively.
Let $A$ be the area of each plate and $d$ be the separation between the plates.
Thus, we have, $\sigma = \dfrac{Q}{A}$ .
The electric field strength between the plates is given by, $E = \dfrac{\sigma }{{{\varepsilon _0}}}$ where $\varepsilon$ represents the absolute permittivity of the dielectric material being used. The dielectric constant, ${\varepsilon _0}$ is also known as the permittivity of free space.
Potential difference between the plates M and N, ${V_{MN}} = Ed = \dfrac{{\sigma d}}{{{\varepsilon _0}}}$ .
Substituting, $\sigma = \dfrac{Q}{A}$ , in the above equation, we have,
${V_{MN}} = \dfrac{{Qd}}{{A{\varepsilon _0}}}$ .
Capacitance is given by, $C = \dfrac{Q}{{{V_{MN}}}} = \dfrac{Q}{{\dfrac{{Qd}}{{A{\varepsilon _0}}}}}$ .
Simplifying, we have, $C = \dfrac{{A{\varepsilon _0}}}{d}$ .
The capacitance of a parallel plate capacitor is proportional to the area, $A\left( {{m^2}} \right)$ of the smallest of the two plates and inversely proportional to the distance or separation, $d$ (i.e. the dielectric thickness) given in metres between these two conductive plates.
Thus, the capacity can be increased by,
-Increasing the area of the plates
-Decreasing the distance between the plates.

Note:
Capacitance is the electrical property of a capacitor and is the measure of a capacitor's ability to store an electrical charge onto its two plates with the unit of capacitance being the Farad (abbreviated to $F$ ) named after the British physicist Michael Faraday. Capacitance is defined as being that a capacitor has the capacitance of One Farad when a charge of One Coulomb is stored on the plates by a voltage of One volt. Note that capacitance, C is always positive in value and has no negative units. However, the Farad is a very large unit of measurement to use on its own so sub-multiples of the Farad are generally used such as micro-farads, nano-farads in pico-farads, for example.