Energy stored in a capacitor

In a nutshell

The work done depositing the charge onto the plates of the capacitor is the same as the energy stored by the capacitor. The area under a $V$ against $Q$ graph gives the energy stored by a capacitor.

Equations

Description	Equation
Capacitance equation	$C=\frac QV$
Energy stored by a capacitor	$W=\frac 12 QV \newline \\[0.1in]W=\frac 12 V^2C \newline \\[0.1in]W=\frac 12 \frac {Q^2}{C} \newline \\[0.1in]$

Variable definitions

Quantity Name	Symbol	Derived Unit	SI BASE Units
$capacitance$	$C$	$F$	$kg^{-1}\,m^{-2}\,s^4\,A^2$
$charge$	$Q$	$C$	$A\,s$
$potential \ difference$	$V$	$V$	$kg\,m^2\,s^{-3}\,A^{-1}$
$enery \ stored \ by \ capacitor$	$W$	$J$	$kg\,m^2\,s^{-2}$

Energy stored by capacitors

When a capacitor is charged, one of the plates becomes negatively charged and the opposite plate becomes positively charged.

The work done depositing electrons on the negative plate and removing the electrons from the positive plate, is supplied by the power supply. This work is then stored as the electric potential between the plates.

The equation for charge stored on a capacitor is given as:

$Q=CV$

The charge stored by a capacitor is directly proportional to the potential difference across it. This means that the graph will be a straight line which passes through the origin:

Physics; Capacitance; KS5 Year 12; Energy stored in a capacitor

A	Potential difference
B	Charge

The work done on the capacitor plate, or the energy stored, is the product of the charge and the average potential difference.

The average potential difference can be calculated as:

$V_{av}=\frac {V_i+V_f }2\newline \\[0.1in]V_{av}=\frac {0+V }2\newline \\[0.1in]V_{av}=\frac {V }2$

Therefore the energy stored by the capacitor is:

$W=Q\times \frac V2 \newline \\[0.1in]W=\frac 12 QV$

It is also the area under a $V$ against $Q$ graph.

Physics; Capacitance; KS5 Year 12; Energy stored in a capacitor

A	Potential difference
B	Charge
1	Energy stored

Equation variations

Using variations of the capacitance equation, two other derivations for the energy stored by a capacitor can be derived:

Using $Q=CV$ to eliminate $Q$ :

$W=\frac 12 (CV)V \newline \\[0.1in]W=\frac 12 V^2C$

Using $V=\frac QC$ to eliminate $V$ :

$W=\frac 12 Q(\frac QC) \newline \\[0.1in]W=\frac 12 \frac {Q^2}C$

Example

A $48 \, \mu F$ capacitor is charged using a $12 \, V$ battery. Calculate the amount of energy stored in the capacitor.

Firstly, write down the known values:

$C=48 \, \mu F = 48 \times 10^{-6} \, F \newline \\[0.1in]V=12 \, V$

Next, write down the equations needed and rearrange if necessary:

$W=\frac 12 V^2C$

Then, substitute the values into the equation:

$W=\frac 12 \times 12^2\times 48\times 10^{-6} \newline \\[0.1in]W=3.456\times 10^{-3}$

Make sure to include units and round to the lowest number of significant figures of the values given by the question:

$energy \ stored \ by \ the \ capacitor, \ W=3.5 \times 10^{-3} \, J$

The energy stored by the capacitor is $\underline{3.5 \times 10^{-3} \, J}.$ 