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SS3: PHYSICS - 1ST TERM

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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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