Back to Course

0% Complete
0/0 Steps

• ## Do you like this content?

Lesson 5, Topic 3
In Progress

# Workdone & Energy Stored in a Capacitor

Lesson Progress
0% Complete

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 Â  Â  Â  Â  Â  Â  Â  Â

W = F x dÂ

Fd = Qv

Therefore Â  F =$$\frac{Qv}{d}$$Â

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

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

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

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

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

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

W =Â $$\frac{1}{2} \frac{q^2}{c}$$

Â But q = cv

W =Â $$\frac{1}{2} \frac{(cv)^2}{c}$$

W = Â $$\frac{1}{2} \scriptsize CV^2$$

Examples

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

Solution

C = 100ÂµF = 1.0 x 10-4 F

V = 16v

W = Â $$\frac{1}{2} \scriptsize CV^2$$

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

= 1.28 x 10-3J

error: