Question

How can capacitance be increased?

Answer 1

Hint: The ability of a system to store the electric charge is known as the capacitance. The capacitance of the parallel plate capacitor depends upon the value of the area of the plates, distance between the plates and the medium between the plates. Hence, to increase the capacitance, one or more of these conditions can be changed.

Complete answer:
A device formed by two conductors insulated from each other is called a capacitor.Capacitance is basically the ability of a system to store the charge and it is technically defined as the ratio of the charge on the conductor/s to the potential difference between the conductors which is mathematically expressed as
$C=\dfrac{Q}{V}$
From this formula, we can derive the capacitance for a parallel plate capacitor which can be expressed as
$C=\dfrac{{{\varepsilon }_{0}}A}{d}$ , where
$C$ = Capacitance
${\varepsilon}_0$ = Permittivity of free space = $8.854\,C^2N^{-1}m^{-2}$
$A$ = Area of the two parallel plates
$d$ = Distance between the parallel plates

Normally, the space between the plates is with nothing but vacuum, and hence we take the permittivity of free space in the equation of capacitance. But as we know, that vacuum is non polarizable and hence transfers relatively less amount of charge. Hence, let us replace the complete space between the plates of the capacitor with a polarizable dielectric material whose permittivity is expressed as $\varepsilon$ .Hence, now the capacitance with dielectric material is expressed as
$C'=\dfrac{\varepsilon A}{d}$

Now, the ratio of the permittivity of dielectric material and permittivity of vacuum is defined as the Dielectric constant of the medium, which is mathematically expressed as
$\dfrac{\varepsilon }{{{\varepsilon }_{0}}}=K$
$\Rightarrow\varepsilon =K{{\varepsilon }_{0}}$
We can substitute this value in the equation of capacitance as,
$C'=\dfrac{K{{\varepsilon }_{0}}A}{d}$
$\therefore C'=KC$
Hence, the capacitance with the dielectric material is $K$ times the capacitance with free space between the plates. It has been experimentally proven that the value of $K$ is always greater than $1$ . Hence, the capacitance with a dielectric material will always be greater than the capacitance with free space between the plates.

Hence, we can increase the capacitance by inserting a dielectric material between the plates.

Note: From the formula of the capacitance, we can see that capacitance is also proportional to the area of the plates, and inversely proportional to the distance between the plates. Hence, the capacitance can also be increased by either increasing the area of the plates or decreasing the distance between the plates. But, we can understand from the equation, that to even increase the capacitance by $1\mu F$ , the area of the plates needs to be increased by several hundred metres or the distance between the plates to be decreased to a few micrometers. Both of the conditions are practically not possible. Hence, we increase the capacitance with the use of dielectric material only.