Lesson 4, Topic 4
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# Potential Difference

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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 = $$\frac{W}{q}$$

Electric field intensity (E) can also be defined as

E =  $$\frac{V}{d}$$

d = distance between two points in an electric field

Therefore, V = Ed _______________ (1)

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

or

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

Substituting this in equation 1

Or V =  $$\frac{Q}{4 \pi \varepsilon_0 d}$$

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

Examples

1. A point charge of magnitude 2 x 10-6C 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 =$$\frac{F}{q}$$          , F=QE

Work done = F x d

= Eq x d

= 30 x 2 x 10-6 0.50

= 3.0 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 = $$\frac{v}{d} = \frac{50}{0.1}$$

=500vm-1

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