When two points A and B are connected in an electric field, more work is done in bringing a unit positive charge from infinity to A than to B, there is potential difference (p.d) between A and B and this difference is equal to the work done in moving the positive charge from B to A against electric field.

Therefore, potential difference between two points is the work done in taking a unit positive charge from one point to the other in the electric field.

The unit of potential difference is volt.

W = qV

V = \( \frac{W}{q} \)Â

Electric field intensity (E) can also be defined as

E =Â \( \frac{V}{d} \)

d = distance between two points in an electric fieldÂ

Â Therefore, V = Ed _______________ (1)

E = \( \frac{Q}{4 \pi \varepsilon_0 r^2} \)

or

V =Â \( \frac{Q}{4 \pi \varepsilon_0 d^2} \scriptsize \; \times \; d\)

Substituting this in equation 1

Or V =Â \( \frac{Q}{4 \pi \varepsilon_0 d} \)

V is the potential due to a charge q at a distance d from the charge.

**Examples**

- A point charge of magnitude 2 x 10
^{-6}C is moved through a distance of 0.50m against a uniform electric field intensity 50vm^{-1}. What is the workdone on the charge?

**Solution**

E =\( \frac{F}{q} \)Â Â Â Â Â , F=QE

Work done = FÂ x d

Â Â Â Â Â Â Â Â Â = EqÂ x d

Â Â Â Â Â Â Â Â = 30 xÂ 2 x 10^{-6} 0.50

Â Â Â Â Â Â Â Â Â = 3.0Â x 10^{-5} J

2. Two parallel plates are charged to a voltage of 50v. If they are separated by a difference of 10.0cm, calculate electric intensity between them.

**Solution**

E =Â \( \frac{v}{d} = \frac{50}{0.1} \)Â

=500vm^{-1}

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