# A-level Physics (Advancing Physics)/Capacitors/Worked Solutions

1. A 2 mF capacitor is connected to a 10V DC power supply. How much charge can be stored by the capacitor?

${\displaystyle Q=CV=2\times 10^{-3}\times 10=0.02{\mbox{ C}}}$

Note that this is the maximum charge - since capacitors charge, as well as discharge, exponentially, we would have to leave the capacitor charging for an infinitely long period of time to charge it to capacity.

2. What is the highest possible energy stored by this capacitor?

${\displaystyle E={\frac {1}{2}}QV=0.5\times 0.02\times 10=0.1{\mbox{ J}}}$

3. The capacitor is placed in series with a 5Ω resistor and charged to capacity. How long would it take for the charge in the capacitor to be reduced to 1 mC?

${\displaystyle Q=Q_{0}e^{-{\frac {t}{RC}}}}$

${\displaystyle Q_{0}=Qe^{\frac {t}{RC}}}$

${\displaystyle e^{\frac {t}{RC}}={\frac {Q_{0}}{Q}}}$

${\displaystyle {\frac {t}{RC}}=\ln {\frac {Q_{0}}{Q}}}$

${\displaystyle t=RC\ln {\frac {Q_{0}}{Q}}=5\times 0.002\times \ln {\frac {0.02}{0.001}}=0.0300{\mbox{ s}}}$

4. After this time has elapsed, how much energy is stored in the capacitor?

${\displaystyle E={\frac {1}{2}}QV={\frac {1}{2}}Q_{0}V_{0}e^{-{\frac {2t}{RC}}}=0.5\times 0.02\times 10\times e^{-{\frac {2\times 0.03}{5\times 0.002}}}=0.25\;m{\mbox{J}}}$

5. What is the capacitance of the equivalent capacitor to the following network of capacitors?

First, ignore the fact that the capacitances are in mF, not F - we will not be using any other units, so if we put in mF, we will get out mF.

Then work out the equivalent capacitance of the second row:

${\displaystyle {\frac {1}{\Sigma C_{2}}}={\frac {1}{1}}+{\frac {1}{2}}+{\frac {1}{3}}={\frac {11}{6}}}$

${\displaystyle \Sigma C_{2}={\frac {6}{11}}{\mbox{ mF}}}$

Then work out the equivalent capacitance of the third row:

${\displaystyle {\frac {1}{\Sigma C_{3}}}={\frac {1}{2}}+{\frac {1}{2}}=1}$

${\displaystyle \Sigma C_{3}=1{\mbox{ mF}}}$

The equivalent capacitance of the first row is easy, since it contains just 1 capacitor: 3mF. So, the total equivalent capacitance is:

${\displaystyle \Sigma C=3+{\frac {6}{11}}+1={\frac {50}{11}}\approx 4.55{\mbox{ mF}}}$