Calculating Capacitor Energy | CAIE 9702 Paper 4 Notes

Paper 4

Physics9702

Calculating Capacitor Energy

Start with W = ½QV. Replace Q or V only when the question does not give it.

Choose the form that matches the data

W=\dfrac{1}{2}QV

Symbols

W=\dfrac{1}{2}CV^2

W=\dfrac{Q^2}{2C}

Given C and V

Use W = ½CV².

Given Q and C

Use W = Q²/(2C).

Worked example

Find p.d. and charge

A 470 μF capacitor stores 19 mJ. Find its p.d. and charge (9702/41/O/N/25 Q6(b)).

V=\sqrt{\dfrac{2W}{C}}

V=\sqrt{\dfrac{2(19\times10^{-3})}{470\times10^{-6}}}

V=9.0\ \mathrm{V}

Q=CV=(470\times10^{-6})(9.0)

Answer

Q=4.2\times10^{-3}\ \mathrm{C}

Common mistake

For an energy change, find final energy minus initial energy. Do not square the change in p.d.

Your turnquick check

A 100 μF capacitor is at 20 V. Find its stored energy.

Show worked answer

W=\dfrac{1}{2}CV^2

W=\dfrac{1}{2}(100\times10^{-6})(20)^2

Answer

W=2.0\times10^{-2}\ \mathrm{J}

Calculating Capacitor Energy

Start with W = ½QV. Replace Q or V only when the question does not give it.

W=\dfrac{1}{2}QV

Symbols

W=\dfrac{1}{2}CV^2

W=\dfrac{Q^2}{2C}

Given C and V

Use W = ½CV².

Given Q and C

Use W = Q²/(2C).

Worked example

Find p.d. and charge

A 470 μF capacitor stores 19 mJ. Find its p.d. and charge (9702/41/O/N/25 Q6(b)).

V=\sqrt{\dfrac{2W}{C}}

V=\sqrt{\dfrac{2(19\times10^{-3})}{470\times10^{-6}}}

V=9.0\ \mathrm{V}

Q=CV=(470\times10^{-6})(9.0)

Answer

Q=4.2\times10^{-3}\ \mathrm{C}

Common mistake

For an energy change, find final energy minus initial energy. Do not square the change in p.d.

Your turnquick check

A 100 μF capacitor is at 20 V. Find its stored energy.

Show worked answer

W=\dfrac{1}{2}CV^2

W=\dfrac{1}{2}(100\times10^{-6})(20)^2

Answer

W=2.0\times10^{-2}\ \mathrm{J}