Question

A 100 pF capacitor is charged to a potential difference of 24 V. It is connected to an uncharged capacitor of capacitance 20 pF. What will be the new potential difference across the 100 pF capacitor ?

Answer 1

Step 1: Given data

Capacitance $C_1=100pF,$

Voltage $V=25V$

New capacitance $C_2=20pF$

Step 2: Formula used

$q=CV$

Step 3: Finding total charge

Total charge $q=C_{1}V=24\times 100pC$

When the charge capacitor is connected with the uncharged capacitor

We can write

\(q_1+q_2=q=24\times 100~qF)

Step 4: Finding the charge on $C_1$

$\frac{q_1}{C_1}=\frac{q_2}{C_2}$ (since voltage is same)

Putting values

$\frac{q_1}{100}=\frac{q_2}{20}$

Or, $q_1=5q_2$

$q_1=\frac{24\times 100qF}{6}$

$q_1=20\times 100pF$

Step 5: Finding the potential at $C_1$

$v_1=\frac{q_1}{C_1}$

$=\frac{20\times 100pF}{100pF}=20V$

Hence the new potential difference is 20 V.