19.1.5 ENERGY STORED IN A CAPACITOR
- Calculating Energy Stored in a Capacitor
- Recall the electric potential energy is the area under a potential-charge graph
- This is equal to the work done in charging the capacitor to a particular potential difference
- The shape of this area is a right angled triangle
- Therefore the work done, or energy stored in a capacitor is defined by the equation:
$
W=\frac{1}{2} Q V
$
- Substituting the charge with the capacitance equation $\mathrm{Q}=\mathrm{CV}$, the work done can also be defined as:
$
\mathrm{W}=\frac{1}{2} \mathrm{CV}^2
$
- Where:
- $\mathrm{W}=$ work done/energy stored $(\mathrm{J})$
- $\mathrm{Q}=$ charge on the capacitor (C)
- $\mathrm{V}=$ potential difference $(\mathrm{V})$
- $C=$ capacitance $(F)$
- By substituting the potential V, the work done can also be defined in terms of just the charge and the capacitance:
$
\mathrm{W}=\frac{Q^2}{2 C}
$
Worked example: Energy stored in a capacitor
Calculate the change in the energy stored in a capacitor of capacitance $1500 \mu \mathrm{F}$ when the potential difference across the capacitor changes from $30 \mathrm{~V}$ to $10 \mathrm{~V}$.
Answer/Explanation
Step 1: Write down the equation for energy stored, in terms of capacitance $C$ and p.d $V$
$
\mathrm{W}=\frac{1}{2} \mathrm{CV}^2
$
Step 2: $\quad$ The change in energy stored is proportional to the change in p.d
$
\Delta \mathrm{W}=\frac{1}{2} \mathrm{C} \Delta \mathrm{V}^2=\frac{1}{2} \mathrm{C}\left(\mathrm{V}_2^2-\mathrm{V}_1^2\right)
$
Step 3:
Substitute in values
$
\Delta \mathrm{W}=\frac{1}{2} \times 1500 \times 10^{-6} \times\left(30^2-10^2\right)=\mathbf{0 . 6} \mathbf{J}
$