Charge and Energy

A capacitor is a device for storing charge. It consists of two plates separated by an insulator.

A graph of the charge stored on the plates against the p.d. across the plates of a capacitor is shown below.

graph of charge against the p.d. across capacitor

${\rm{Capacitance}}\,{\rm{of}}\,{\rm{a}}\,{\rm{capacitor = }}{{{\rm{charge}}\,{\rm{stored}}\,{\rm{on}}\,{\rm{the}}\,{\rm{plates}}} \over {{\rm{potential}}\,{\rm{difference}}\,{\rm{across}}\,{\rm{the}}\,{\rm{plates}}}}$

or $C = {Q \over V}$

The unit for capacitance is the farad (F). One farad is equal to one coulomb per volt.

1 F = 1 C V^-1.

Capacitors normally have very small values and so are often measured in mF (10^-3 F), $\mu$ F (10^-6 F), nF (10^-9), or pF (10^-12 F)

Electrical energy is required to move charges onto the plates of a capacitor. This energy is stored in the capacitor and is the result of work done in transferring the charge from the supply to the plates.

Energy stored in capacitor

= work done in charging capacitor

= area under charge against voltage graph for a capacitor

= ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$ QV

= ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$ CV²

= ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}{{Q^2 } \over C}$

Question: A capacitor stores 2 mC of charge when the p.d. across the capacitor plates is 100 V. What is the capacitance of the capacitor?

Answer

C = ?

Q = 2 mC = 2 × 10^-3 C

V = 100 V

$C = {Q \over V} = {{2 \times 10^{ - 3} } \over {100}} = 2 \times 10^{ - 5} F$

Question: Calculate the energy stored in a 10 $\mu$ capacitor when the p.d. across the plates is 20 V.

Answer

E = ?
C = 10 $\mu$ F
Q = ?
V = 20 V

E = ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$ QV = ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$ CV²
E = ${\raise0.5ex\hbox{$\scriptstyle 1$} \kern-0.1em/\kern-0.15em \lower0.25ex\hbox{$\scriptstyle 2$}}$ × 10 × 10^-6 × 20²
E = 0.002 J