Question

A potential difference of 75V is applied to a combination of 1.25microFarad and 0.6microfarad capacitor connected in series (a) what is the charge on each capacitor? (b) what is the potential difference across the 1.25microFaradcapacitor?

Answer 1

(a) For the series connection of capacitors, the magnitude of charge on each capacitor will be the same:

$$Q_1=Q_2=Q.$$

So, let's first find the equivalent capacitance of two capacitors connected in series:

$$C_{eq}=\dfrac{C_1C_2}{C_1+C_2},$$ $$C_{eq}=\dfrac{1.25\cdot10^{-6}\ F\cdot0.6\cdot10^{-6}\ F}{1.25\cdot10^{-6}\ F+0.6\cdot10^{-6}\ F}=0.405\ \mu F.$$

Finally, we can find the charge on each capacitor:

$$Q=Q_1=Q_2=C_{eq}\Delta V,$$ $$Q=0.405\cdot10^{-6}\ F\cdot75\ V=30.4\ \mu C.$$

(b) We can find the potential difference across $1.25\ \mu F$ capacitor as follows:

$$\Delta V=\dfrac{Q}{C_1}=\dfrac{30.4\cdot10^{-6}\ C}{1.25\cdot10^{-6}\ F}=24.32\ V.$$