Question

Two capacitors of 3pf and 6pf are connected in series and a potential difference of 5000v is applied across the combination. They are then disconnected and reconnected in parallel. The potential between the plates is

Answer 1

Given:

The capacitance of both the capacitors is $C_{1}=3pF$ and $C_{2}=6pF$ respectively.

The potential difference across the circuit is $V=5000volts.$

To Find:

The potential between the plates is

Explanation:

Case 1: when the capacitors are connected in series, net capacitance is:

$\frac{1}{C_{eq.}}=\frac{1}{C_{1}} +\frac{1}{C_{2}} \\\frac{1}{C_{eq.}}=\frac{1}{3} +\frac{1}{6}\\{C_{eq.}}=2pF$

We know that $Q=C \times V$:

$Q=C \times V\\Q=(2 \times 10^{-12}) \times 5000\\Q=10^{-8}C$

Case 2: when the capacitors are connected in parallel, net capacitance is:

$C_{eq.}=C_{1}+C_{2}\\C_{eq.}=3+6\\C_{eq.}=9pF$

It is very clear that in the parallel case, the voltage remains the same at every point, thus:

$Q_{1}+Q_{2}=2Q\\C_{1}V+C_{2}V=2Q\\V(C_{1}+C_{2})=2Q\\V \times (3 \times 10^{-12}+6 \times 10^{-12})=2 \times 10^{-8}\\V=2222volts$

Hence, the potential difference across the plates is calculated to be $2222volts.$