# Capacitor And Capacitance

## Capacitor and Capacitance

### Introduction

The first question that comes to our mind when we hear about capacitors is – ‘What is a capacitor?’ We know what a resistor is; it allows the current to pass through it and dissipate heat but can’t store electrical energy. So there was a need to develop a device that can store electrical energy.The most common arrangement consists of a set of conductor’s also known as conducting plates having charges on them and separated by an insulator.

The conducting plates have some charges q1 and q2 (Usually if one plate has +q the other has –q charge).The electric field in the region between the plates depends on the charge given to the conducting plates. We also know that potential difference (V) is directly proportional to electric field hence we can say,

Q ∝ V

Q = CV

C = $\frac Qv$

This constant of proportionality is known as capacitance of capacitor. From the equation it may seem that ‘C’ depends on charge and voltage, but actually it depends on shape and size of the capacitor and also on the insulator used between the conducting plates. Its SI unit is Farad.

Parallel Plate Capacitor

This is one of the most common types of capacitor used. It has two parallel plates separated by a distance d (as shown in the figure).

From Gauss law we know that electric field due to a plane charged sheet is

E = $\frac {D}{2\epsilon }$

Where, D is the charge density on the sheet.

Suppose plate P1 is positively charged and P2 is negatively charged. Then Electric field in the region to the left of P1,

E =   $\frac {D}{2\epsilon}$   –   $\frac {D}{2 \epsilon}$

E = 0

Similarly it is zero for the region to the right of P2.In between the plates the electric field will be,

E = $\frac {D}{2 \epsilon}$      +   $\frac {D}{2 \epsilon}$

E =   $\frac {D}{2 \epsilon}$

The direction of this field is from positive to negative plate. Now we will try to find the expression for capacitance.

V = Ed

V = $\frac {Qd}{A\epsilon}$

We also know, C = $\frac Qv$

From the equations we get,

C = $\frac {A\epsilon}{d}$

As we can see from the above equation it only depends on the dimensions of the conducting plates.

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#### Practise This Question

A potential difference of 600 V is applied across the plates of a parallel plate capacitor. The separation between the plates is 3 mm. An electron projected vertically, parallel to the plates, with a velocity of 2×106 ms1 moves undeflected between the plates. What is the magnitude of the magnetic field between the capacitor plates?