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
- 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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