Name: Capacitance of a Parallel Plate Capacitor
Uploaded: 2019-07-19
Duration: 4 min 13 s
Description: Capacitors are available in various shapes and sizes. In this video, we will learn how to calculate the capacitance of a parallel plate capacitor.

Question

Derive an expression for the capacitance of a parallel-plate capacitor in terms of the areas of the plates and the distance between them. Assume there is a vacuum in between the plates with the electric field between the plates

E = σ / ε_{0} .

Accepted Answer

A parallel-plate capacitor consists of two long, plane, metallic plates mounted on two insulating stands and placed at a small distance apart. The plates are exactly parallel to each other and the space between them is filled with air or some dielectric material.
In Fig., $X$ and $Y$ Are two plane metallic plates, each of area A ${metre}^{2}$ And having a distance of d metre between them. The space between the plates is completely filled with a dielectric material of dielectric constant $K$ .

Suppose the plate $X$ Is given a charge of $+ Q$ Coulomb. By induction, $- Q$ Coulomb of charge is produced on the inner surface of the plate $Y$ and $+ Q$ Coulomb on the outer surface. Since, the plate $Y$ Is connected to the earth, the $+ Q$ Charge on the outer surface flows to the earth. Thus, the plates $X$ and $Y$ Have equal and opposite charges. All the lines of force starting from the plate $X$ Reach the plate $Y$ And except near the edges, the electric field between the plates is uniform everywhere.

Suppose, the surface density of charge on each plate is $σ$ . We know that the intensity of electric field at a point between two plane parallel sheets of equal and opposite charges is $σ / ε_{0}$ , where $ε_{0}$ Is the permittivity of free space. If the space between the sheets is filled with a dielectric material of dielectric constant $K$ , then intensity of electric field will be given by

$E = \frac{σ}{K ε_{0}}$

The charge on each plate is $Q$ And the area of each plate is $A$ . Thus, $σ = Q / A$ , and so

$E = \frac{Q}{{Kε}_{0} A} . . . . (i)$

Now, let the potential difference between the two plates be $V$ Volt. Then the electric field between the plates is given by.

$E = V / d$

or $V = E d .$

Substituting the value of $E$ From Equation $(i)$ , we get $V = \frac{Qd}{{Kε}_{0} A}$

So, capacitance of the capacitor is

$C = \frac{Q}{V} = \frac{Q}{Qd / {Kε}_{0} A}$

or $C = \frac{{Kε}_{0} A}{d} farad, . . . (ii)$

where $ε_{0} = 8.85 x 10^{- 12} farad / metre$ . It is clear from this formula that in order to obtain high capacitance, $(i)$ A should be large, that is, the plates of large area should be taken, $(ii)$ D should be small, that is, the plates should be kept close to each other, and $(iii)$ Space between the plates should be filled with a dielectric of high dielectric constant $K$ .
If there is vacuum (or air) between the plates, then $K = 1$ And capacitance of the parallel plate (air) capacitor is

$C_{0} = \frac{ε_{0} A}{d} farad .$

Derive an expression for the capacitance of a parallel-plate capacitor in terms of the areas of the plates and the distance between them. Assume there is a vacuum in between the plates with the electric field between the plates $E = σ / ε_{0} .$

Important Questions on Capacitors and Dielectrics

Learn and Prepare for any exam you want !

Derive an expression for the capacitance of a parallel-plate capacitor in terms of the areas of the plates and the distance between them. Assume there is a vacuum in between the plates with the electric field between the plates E=σ/ε0 .

Important Questions on Capacitors and Dielectrics

Learn and Prepare for any exam you want !

Derive an expression for the capacitance of a parallel-plate capacitor in terms of the areas of the plates and the distance between them. Assume there is a vacuum in between the plates with the electric field between the plates $E = σ / ε_{0} .$