__What is a Parallel plate Capacitor?__

*Parallel plate capacitor is defined as a combination of two conducting parallel surfaces separated by a non conducting medium such that it can store electric charge.*

- When the plates of capacitor are connected to the two terminals of a battery, the surfaces get charged by charges of equal magnitude but opposite sign.
- When the battery is disconnected, the charges on the conducting surfaces remains the same.
- Hence, a parallel plate capacitor is used for storing the
*charge*.

### Capacitance of Parallel plate Capacitor

Consider about a parallel plate capacitor consisting of plates ( R ) \ \text {and} \ ( S ) separated by a distance ( d ) as shown in figure.

Let, ( + \sigma ) \ \text {and} \ ( - \sigma ) are the *surface charge density* of plates ( R ) \ \text {and} \ ( S ) respectively.

Then *electric field* between the plates will be –

E = \left ( \frac {\sigma}{\epsilon_0} \right ) ……… (1)

Here, ( \epsilon_0 ) is the *absolute permittivity* of free space or vacuum or air.

Let, ( V ) is the potential difference applied between the plates of capacitor.

Then, magnitude of *electric field* in terms of potential will be –

E = \left ( \frac {dV}{dr} \right ) = \left ( \frac {V}{d} \right )

So, \quad V = E d …….. (2)

Using equation (1) and (2), we will get –

V = \left ( \frac {\sigma }{\epsilon_0} \right ) d

= \left ( \frac {q}{\epsilon_0 A} \right ) d ………. (3)

Because \quad \sigma = \left ( \frac {q}{A} \right )

From definition of *capacitance*, we will get –

C = \left ( \frac {q}{V} \right )

= \left [ \frac {q}{\left ( \frac {q}{\epsilon_0 A} \right ) d} \right ]

= \left [ \frac {q \epsilon_0 A}{qd} \right ]

Therefore, \quad C = \left ( \frac {\epsilon_0 A}{d} \right )

*Thus capacitance of a parallel plate capacitor is ( i ) directly proportional to the area of plates and ( ii ) inversely proportional to distance between plates.*

**TO BE NOTED –**

**Capacitance depends on –** (1) The size i.e. area ( A ) of plates of capacitor and (2) Distance ( d ) between the plates. Thus, capacitance is determined by geometry of plates. It is a design parameter.

**Capacitance doesn’t depends on –** (1) The charge ( q ) on the plates. (2) Applied potential difference ( V ) between the plates.

__Working principle of a Capacitor__

Consider that a thin positively charged conducting plate A is kept in air. Let, another uncharged conducting plate B of similar size and shape is brought nearer to the plate A as shown in figure (A).

Then, *negative charge* will induce in nearer face of plate B and *positive charge* on farther face of plate B as shown in figure (B).

The induced negative charge on plate B will tend to decrease the *potential* of plate A ( because positive and negative charges will cancel to each other) where as induced positive charge will tend to increase the potential of plate B .

Since, induced negative charge on plate B is nearer to plate A in compared to the induced positive charge on plate B , therefore there will be a net decrease in the potential of plate A .

In order to keep the potential of plate A same as before, more charge can be supplied to it.

*Thus the capacity of charged conducting plate A has increased by bringing an uncharged conducting plate B nearer to it.*

If, farther side face of plate B is now connected to the earth, then all induced positive charges on plate B will drain to earth as shown in figure (C). Now, the plate B will have only induced negative charges.

This activity will further reduce the potential of plate A and thus more charges can be supplied to it for this. Thus a huge amount of charge can be stored in this arrangement.

*In this way, a capacitor works to store huge amount of charge.*

__Effect of Dielectric on Capacitance__

Capacitance of a parallel plate capacitor with free space between the plates is –

C_0 = \left ( \frac {\epsilon_0 A}{d} \right ) ……… (1)

Due to charged plates of capacitor, an electric field ( \vec {E_0} ) will set up between the plates. Now, let a *dielectric* slab of thickness ( t ) is inserted between the plates as shown in figure.

Due to *polarization of dielectric*, equal and opposite *induced charges* will appear on the two surfaces of the slab. These induced charges give raise to *induced electric field* ( \vec {E_p} ) which is in a direction opposite to the direction of applied field ( \vec {E_0} ) .

Now, the reduced value of electric field in the *dielectric* is given by [ E = ( E_0 - E_p ) ] which exist in the region of thickness ( t ) and ( E_0 ) field exist in the region of thickness ( d - t ) .

Then, potential difference between the two plates of the capacitor will be –

V = E_0 \left ( d - t \right ) + E ( t )

= E_0 \left ( d - t \right ) + \left ( \frac {E_0}{K} \right ) ( t )

= E_0 \left [ \left ( d - t \right ) + \left ( \frac {t}{K} \right ) \right ] ……….. (2)

From definition of *dielectric constant*, we get –

K = \left ( \frac {E_0}{E} \right )

Also, \quad E_0 = \left ( \frac {\sigma}{\epsilon_0} \right ) = \left ( \frac {q}{A \epsilon_0} \right )

Then, equation (2) becomes –

V = \left ( \frac {q}{A \epsilon_0} \right ) \left [ \left ( d - t \right ) + \left ( \frac {t}{K} \right ) \right ] ……….. (3)

Therefore capacitance of the capacitor will be –

C = \left ( \frac {q}{V} \right ) = \left ( \frac {q A \epsilon_0}{q \left [ ( d - t ) + \left ( \frac {t}{K} \right ) \right ]} \right )

= \left ( \frac {\epsilon_0 A}{ \left ( d - t \right ) + \left ( \frac {t}{K} \right )} \right )

= \left ( \frac {\epsilon_0 A}{d \left [ \left ( 1 - \frac {t}{d} \right ) + \left ( \frac {t}{dK} \right ) \right ]} \right ) ………. (4)

Dividing equation (4) by equation (1), we get –

\left ( \frac {C}{C_0} \right ) = \left ( \frac {\epsilon_0 A}{d} \right ) \div \left ( \frac {\epsilon_0 A}{d \left [ \left ( 1 - \frac {t}{d} \right ) + \left ( \frac {t}{dK} \right ) \right ]} \right )

= \left [ \frac {1}{{\left [ \left ( 1 - \frac {t}{d} \right ) + \frac {t}{dK} \right ]}} \right ]

= \left [ \frac {1}{1 - \frac {t}{d} \left ( 1 - \frac {1}{K} \right ) } \right ]

So, \quad C = \left [ \frac {C_0}{1 - \frac {t}{d} \left ( 1 - \frac {1}{K} \right ) } \right ] ………. (5)

From equation (5), it is clear that ( C > C_0 )

*Thus, capacitance of a capacitor increases by introducing a dielectric slab between the plates of a capacitor.*

### Effect of thickness of Dielectric slab

When ( t = d ) i.e. if thickness of dielectric slab is equal to the spacing between the plates then from equation (5), we have –

C = \left ( \frac {C_0}{\frac{1}{K}} \right ) = K C_0

*Thus capacitance of a capacitor increases by a factor ( K ) .*

When, \left ( t = \frac {d}{2} \right ) i.e. if thickness of dielectric slab is equal to the half of the spacing between the plates then from equation (5) we have –

C = \left ( \frac {2 K C_0}{K + 1} \right )

= \left ( \frac {2 K}{K + 1} \right ) C_0

When, \left ( t = \frac {2d}{3} \right ) i.e. if thickness of dielectric slab is equal to the spacing between the plates then from equation (5) we have –

C = \left ( \frac {3 K C_0}{K + 2} \right )

= \left ( \frac {3 K}{K + 2} \right ) C_0

__Induced surface Charge density__

*Surface charge density developed in the dielectric medium between charged parallel plates is known as induced surface charge density.*

Reduced value of electric field between two oppositely charged parallel plates is –

E = ( E_0 - E_p ) ……. (1)

But, \quad E_0 = \left ( \frac {\sigma}{\epsilon_0} \right )

And, \quad E_p = \left ( \frac {\sigma_p}{\epsilon_0} \right )

Also, \quad E = \left ( \frac {E_0}{K} \right ) = \left ( \frac {\sigma}{\epsilon_0 K} \right )

Therefore, equation (1) becomes –

\left ( \frac {\sigma}{\epsilon_0 K} \right ) = \left [ \frac {(\sigma - \sigma_p)}{\epsilon_0} \right ]

Or, \quad \sigma_p = \left ( \sigma - \frac {\sigma}{K} \right ) = \left ( \frac {K - 1}{K} \right ) \sigma

Since, \quad ( K > 1 ) so ( \sigma_p < \sigma )

*Therefore, induced surface charge density in the dielectric is less than the surface charge density of plates.*

**SPECIAL CASE**

If gap between the charged parallel plates is filled with a conducting material instead of a dielectric then –

( E = 0 )

Hence, equation (1) becomes –

( E_p = E_0 )

Therefore, \quad \left ( \frac {\sigma_p}{\epsilon_0} \right ) = \left ( \frac {\sigma}{\epsilon_0} \right )

Or, \quad \sigma_p = \sigma

*Thus, induced surface charge density is equal to the surface charge density of plates.*

See numerical problems based on this article.