Energy Stored in a Capacitor (Cambridge International A-Level Physics)

Calculating Energy Stored in a Capacitor

• Recall that the electric potential energy is equal to the area under a potential-charge graph

  • This is equal to the work done in charging the capacitor across a particular potential difference

• Therefore the work done, or energy stored in a capacitor is defined by the equation:

$W=\frac{1}{2}QV$

• If the charge Q is substituted using the capacitance equation, Q = CV, the work done can also be defined as:

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

• Where:

  • W = work done/energy stored (J)
  • Q = charge on the capacitor (C)
  • V = potential difference (V)
  • C = capacitance (F)

• If the potential V is substituted, the work done can also be defined in terms of the charge and the capacitance:

$W=\frac{Q^{2}}{2C}$

Worked Example Calculate the change in the energy stored in a capacitor of capacitance 1500 μF when the potential difference across the capacitor changes from 10 V to 30 V.

Answer:

Step 1: Write down the equation for energy stored in terms of capacitance C and p.d V

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

Step 2: The change in energy stored is proportional to the change in p.d

$\Delta W = 1 \frac{1}{2} C V_{2}^{2} - 1 \frac{1}{2} C V_{1}^{2} = 1 \frac{1}{2} C ( V_{2}^{2} - V_{1}^{2} )$

Step 3: Substitute in values

$\Delta W = 1 \frac{1}{2} \times ( 1500 \times 10^{- 6} ) \times ( 30^{2} - 10^{2} )$

$\Delta W=0.6J$