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

Hint: 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.