Skip to content
Related Articles

Related Articles

Improve Article

Energy stored in a Capacitor

  • Last Updated : 30 Jun, 2021
Geek Week

Capacitors are used in almost every electronic device around us. From a fan to a chip, there are lots of capacitors of different sizes around us. Theoretically, the basic function of the capacitor is to store energy. Its common usage includes energy storage, voltage spike protection, and signal filtering. It was invented by a German scientist, Ewald George von Kleist, in 1745. Physically, a capacitor is just two conductors which are separated by an insulator. They are able to hold a charge which gives rise to a lot of their properties. Let’s study these properties in detail. 

Capacitors and Capacitance

A capacitor is a system of two charges that are separated by an insulator. Let’s say the two conductors have a charge of Q1 and Q2 and potential V1 and V2.  Usually, the charges are Q and -Q. The electric field in this region is proportional to the charge on the surface of capacitors. The figure below shows two conductors with charges Q and -Q on them. These conductors are separated by an insulator. 

Now it is known that potential is nothing but the work done to bring the charge from the infinite to the present position. In this way, the potential also becomes proportional to the charge on the conductors. This means that the ratio of charge and potential is constant. The ratio is termed capacitance. 

C = \frac{Q}{V}



Although any shape and size works for making a capacitor, in real life most of these capacitors are cylindrical. While doing calculations and for the symbol, the capacitor is denoted as a parallel plate capacitor. 

Parallel Plate Capacitor

A parallel plate capacitor consists of two large parallel planes separated by a small distance. Most of the time, the medium between the plates is considered to be a vacuum, but it can be any insulator material. The figure shows the diagram of a parallel plate capacitor that is connected to a battery. 

Energy stored in a Capacitor

When a battery is connected across the plates of a capacitor, the current charges the capacitor, leading to the accumulation of the charges on the opposite plates. As the charges accumulate, the potential difference between the plates starts increasing. Let’s say the capacitance of the capacitor is “C”, it was uncharged initially. Now, after being connected to the battery, let’s say there is a potential difference, “V” that develops between the plates. Consider “q” as the charge on the plates at that time. Then, 

 q = CV

Work done to bring the charge from infinity to the potential of “V” is given by, 

W = qV

Suppose the battery delivers a charge of “dq” to the plates while keeping the potential constant. Then, 



W = dq.V = \frac{q}{C}dq

In this way, a total charge “q” is transferred from the battery to the capacitor. Then, the work done will be given by

 W = \int \frac{q}{C}dq = \frac{q^2}{2C}

So, the energy stored in the capacitor is, 

E = \frac{q^2}{2C}

Using the previous relationship, q = CV

Energy can be re-written as, 

 E= \frac{C^2V^2}{2C}\\E=\frac{1}{2}CV^2

Using relationship, C= q/V

Energy can be written as,



E= \frac{1}{2}CV^2\\E= \frac{1}{2}\frac{q}{V}V^2\\E=\frac{1}{2}qV^2

Therefore, the three formulae obtained for Energy stored in a capacitor,

 E = \frac{q^2}{2C} = \frac{1}{2}CV^2 = \frac{1}{2}qV

Sample Problems

Question 1: Find the capacitance if a charge of 3C and potential of 9V is maintained on plates. 

Solution: 

The relation for capacitance is given by, 

q = CV

Given: q = 3C and V = 9V

q = CV 

⇒ 3 = C(9) 



⇒ \frac{1}{3} = C 

C = 0.333F

Question 2: Find the capacitance if a charge of 18C and the potential of 3V is maintained between the plates. 

Solution: 

The relation for capacitance is given by, 

q = CV

Given: q = 18C and V = 3V

q = CV 

⇒ 18 = C(3) 

⇒ 6 = C 

C = 6F

Question 3: Find the capacitance if a charge of 1C and the potential of 2V is maintained between plates. 

Solution: 

The relation for capacitance is given by, 

q = CV

Given: q = 1C and V = 2V

q = CV 

⇒ 1 = C(2) 

⇒ 0.5 = C 

C = 0.5F

Question 4: Find the energy stored in the capacitor of 12pF which is connected to a battery of 10V. 

Solution: 

The relation for energy stored in a capacitor is given by,

E = \frac{1}{2}CV^2

Given: C = 12pF and V = 10V

E = \frac{1}{2}CV^2 \\ = E = \frac{1}{2}(12 \times 10^{-12})(10)^2 \\ = E = \frac{1}{2}(12 \times 10^{-12} \times 100) \\ = E = \frac{1}{2}(12 \times 10^{-10}) \\ = E = 6 \times 10^{-10} J

Question 5: Find the energy stored in the capacitor which has a charge of 9 x 10-5C and is connected to a battery of 10V. 

Solution: 

The relation for energy stored in a capacitor is given by,

E = \frac{1}{2}QV



Given: q = 9 x 10-5 and V = 10V

E = \frac{1}{2}QV \\ = E = \frac{1}{2}(9 \times 10^{-5})(10) \\ = E = \frac{1}{2}(9 \times 10^{-4})\\ = E = \frac{1}{2}(9 \times 10^{-4}) \\ = E = 4.5 \times 10^{-4} J

Question 6: Find the energy stored in the capacitor of 3pF which has a charge of 9 x 10-5C.  

Solution: 

The relation for energy stored in a capacitor is given by,

E = \frac{q^2}{2C}

Given: q = 9 x 10-5 and C = 3 x 10-12

E = \frac{q^2}{2C} \\ = E = \frac{(9 \times 10^{-5})^2}{2 \times 3 \times 10^{-12}} \\ = E = \frac{81 \times 100}{6} \\ = E = 1350 J

Question 7: A capacitor of capacitance “C”, is first connected to a battery of voltage V and fully charged, then it is connected to a battery of voltage 2V. Find the electrostatic energy gained/lost in this process. 

Solution: 



First, the capacitor is connected to a battery of the potential V. 

Energy of the capacitor after full charge, 

E1\frac{1}{2}CV^2

After it is connected to another battery, 

E2\frac{1}{2}C(2V)^2

⇒E2\frac{1}{2}4CV^2

Energy change in this process = E2 – E1

                                                               \frac{1}{2}4CV^2 - \frac{1}{2}CV^2

                                               = \frac{3}{2}CV^2

Since, the energy change is positive. Electrostatic energy was gained in the process. 

Attention reader! All those who say programming isn’t for kids, just haven’t met the right mentors yet. Join the  Demo Class for First Step to Coding Course, specifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.




My Personal Notes arrow_drop_up
Recommended Articles
Page :