Question

Capacitors C1=10μF and C2=20μF are each charged to 14 V , then disconnected from the battery without changing the charge on the capacitor plates. The two capacitors are then connected in parallel, with the positive plate of C1connected to the negative plate of C2 and vice versa.
(a) Afterward, what is the charge on each capacitor?
(b) What is the potential difference across each capacitor?

Answer 1

a) $Q_1 = 0, Q_2 = 140 \mu C$

b) $V_1=0, V_2 = 7V$

Explanation:

a)

The charge on each capacitor is given by

$Q=CV$

where

C is the capacitance

V is the voltage

For the 1st capacitor, we have

$Q_1 = C_1 V = (10 \mu F)(14 V)=140 \mu C$

For the 2nd capacitor, we have

$Q_2 = C_2 V=(20 \mu F)(14 V)=280 \mu C$

After being charged, the two capacitors are disconnected from the battery, then they are connected in parallel, with the positive plate of C1 connected to the negative plate of C2. This means that charges will start to flow from one capacitor to the other (due to the opposite polarity), until the net charge on the plate of the capacitor with lower capacitance becomes zero.

Here, this means that $140 \mu C$ of charge will flow from C2 to C1: as a result, the final charge on the capacitor 1 will be

$Q_1 = 0$

While the final charge on capacitor 2 will be

$Q_2 = 280 - 140 = 140 \mu C$

b)

In order to find the potential difference across each capacitor, we use the equation

$V=\frac{Q}{C}$

For capacitor 1, the charge is zero, so

$V_1 = 0$

For capacitor 2, the charge is

$Q_2 = 140 \mu C$

The capacitance is

$C=20 \mu F$

So, the potential difference across it will be

$V_2 = \frac{Q_2}{C_2}=\frac{140}{20}=7 V$