Question

Capacitors C1 = 6.00 x 10^(-6)F and C2 = 2.00 x 10^(-6)F are charged as a parallel combination across a 250V battery. The capacitors are disconnected from the battery and from each other. They are then connected positive plate to negative plate and negative plate to positive plate. Calculate the resulting charge on each capacitor.

Answer 1

Let's first find the charge on each capacitors which they acquire when charged as a parallel combination across a 250 V battery:

$$Q_1=C_1V=6.0\cdot10^{-6}\ F\cdot250\ V=1.5\cdot10^{-3}\ C,$$ $$Q_2=C_2V=2.0\cdot10^{-6}\ F\cdot250\ V=0.5\cdot10^{-3}\ C.$$

Then, the two capacitors are connected positive plate to negative plate and negative plate to positive plate. Let's find the net charge on the circuit:

$$Q_{net}=Q_1-Q_2=1.5\cdot10^{-3}\ C-0.5\cdot10^{-3}\ C=1.0\cdot10^{-3}\ C.$$

Let's find the equivalent capacitance of the parallel combination of two capacitors:

$$C_{eq}=C_1+C_2=6.0\cdot10^{-6}\ F+2.0\cdot10^{-6}\ F=8.0\cdot10^{-6}\ F.$$

Then, we can find the new voltage across each capacitor (the voltage across each element is the same in parallel circuit):

$$V_{new}=\dfrac{Q_{net}}{C_{eq}}=\dfrac{1.0\cdot10^{-3}\ C}{8.0\cdot10^{-6}\ F}=125\ V.$$

Finally, we can find the new charge on each capacitor:

$$Q_{1,new}=C_1V_{new}=6.0\cdot10^{-6}\ F\cdot125\ V=750\ \mu C,$$ $$Q_{2,new}=C_2V_{new}=2.0\cdot10^{-6}\ F\cdot125\ V=250\ \mu C.$$