# Physics Lesson 14.7.4 - Energy Stored in a Charged Capacitor

Welcome to our Physics lesson on Energy Stored in a Charged Capacitor, this is the fourth lesson of our suite of physics lessons covering the topic of Capacitance and Capacitors, you can find links to the other lessons within this tutorial and access additional physics learning resources below this lesson.

## Energy Stored in a Charged Capacitor

The source (here the battery) does some work to charge a capacitor from 0 to Q (i.e. increasing the charge of capacitor by ΔQ). As discussed in the previous tutorials of this section, the work W needed for such a process is

W = ∆Q × ∆V

This process brings a uniform increase of the charge as shown in the graph below.

It is clear that the area under the graph represents the work done by the source to charge the capacitor. Since the figure under the graph is a right triangle, whose area is A = a × b / 2, we obtain for the work done by the source to charge the capacitor

W = Q × ∆V/2

where Q is the charge when the capacitor is fully charged.

As stated earlier in this chapter, the work done on a system contributes in the increase of its potential energy. This is also true in the specific case, i.e. the work done by the source to charge the capacitor from 0 to Q brings an increase in the potential energy of capacitor from 0 to PE. Therefore, we obtain for the electric potential energy stored in a charged capacitor:

W = EPEC = Q × ∆V/2

We can transform the above equation into other forms using the capacitor formula C = Q /ΔV. Thus, we can write

EPEC = Q × ∆V/2
= (C × ∆V) × ∆V/2
= C × ∆V2/2

and

EPEC = Q × ∆V/2
= Q × Q/C/2
= Q2/2C

Obviously, all these formulae are equivalent as they are obtained through equivalent transformations.

### Example 8

Two capacitors C1 = 60 nF and 30 nF are connected to the terminals of a 12 V battery as shown in the figure below.

1. What is the energy stored in the system?
2. What is the energy stored in the capacitor C1?
3. ### Solution 8

a) The total capacitance of this system of capacitors connected in series is

1/Cs = 1/C1 + 1/C2
= 1/60 nF + 1/30 nF
= 1 + 2/60
= 3/60

Thus,

Cs = 60/3 nF
= 20 nF

Therefore, the energy stored in the system is

EPEC == C × ∆V2/2
= 20 nF × (12 V)2/2
= 1440 nJ
= 1.44 μJ
= 1.44 × 10-6 J

b) The capacitor C1 carries the same charge as the entire system as it is connected in series with C2. But first, we must find this charge using the capacitor formula. We have

Q1 = Q2 = Q
= Cs × ∆V
= 20 nF × 12 V
= 240 nC
= 2.4 × 10-7 C

Therefore, we obtain for the energy stored in the capacitor C1 (we have C1 = 60 nF = 6 × 10-8 F):

W1 = EPEC1 = Q2/2C1
= (2.4 × 10-7 C)2/2 × 6 × 10-8 F
= 0.48 × 10-6 J
= 4.8 × 10-7 J

You have reached the end of Physics lesson 14.7.4 Energy Stored in a Charged Capacitor. There are 4 lessons in this physics tutorial covering Capacitance and Capacitors, you can access all the lessons from this tutorial below.

## More Capacitance and Capacitors Lessons and Learning Resources

Electrostatics Learning Material
Tutorial IDPhysics Tutorial TitleTutorialVideo
Tutorial
Revision
Notes
Revision
Questions
14.7Capacitance and Capacitors
Lesson IDPhysics Lesson TitleLessonVideo
Lesson
14.7.1Capacitance
14.7.2Capacitors
14.7.3Combination of Capacitors
14.7.4Energy Stored in a Charged Capacitor

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