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Lesson 5, Topic 3
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# Workdone & Energy Stored in a Capacitor

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Workdone in a Capacitor

The work done in a capacitor is the workforce required to move a unit positive charge from infinity to that point.

W = QV

W = F x d

Fd = Qv

Therefore   F =$$\frac{Qv}{d}$$

But E = $$\frac{Qv}{d} \; \times \; \frac{1}{Q}$$

E = $$\frac{v}{d}$$

Energy Stored in a Capacitor

When charging a capacitor, the average value of the potential difference is given by:

$$\frac{0 + v}{2}$$

= $$\frac{1}{2} \scriptsize v$$

The workdone in charging the capacitor through an average potential difference = $$\frac{1}{2} \scriptsize v$$ is

W= $$\frac{1}{2} \scriptsize v \; \times \; q$$

W =  $$\frac{1}{2} \scriptsize qv$$  ___________(1)

But c = $$\frac{q}{v}$$  or v = $$\frac{q}{c}$$

W =  $$\frac{1}{2} \scriptsize qv = \normalsize \frac{1}{2} \scriptsize q \; \times \normalsize \frac{q}{c}$$

W = $$\frac{1}{2} \frac{q^2}{c}$$

But q = cv

W = $$\frac{1}{2} \frac{(cv)^2}{c}$$

W =  $$\frac{1}{2} \scriptsize CV^2$$

Examples

A capacitor of 100µF is charged to a voltage of 16v. What is the energy stored in the capacitor.

Solution

C = 100µF = 1.0 x 10-4 F

V = 16v

W =  $$\frac{1}{2} \scriptsize CV^2$$

=$$\frac{1}{2} \scriptsize \; \times \; 1.0 \times 10^{-4} \; \times \; 16^2$$

= 1.28 x 10-3J

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