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SS3: PHYSICS - 1ST TERM

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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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