Question

Two capacitors of capacitance $2\, \mu F$ and $3\, \mu F$ are joined in series. Outer plate first capacitor is at $1000$ volt and outer plate of second capacitor is earthed (grounded). Now the potential on inner plate of each capacitor will be

• A. 700 V
• B. 200 V
• C. 600 V
• D. 400 V

Answer 1

Answer: (D)

Equivalent capacitance
$=\frac{2 \times 3}{2+3}=\frac{6}{5} \mu$
Total charge, $Q=C V=\frac{6}{5} \times 1000$
$=1200\, \mu C$
Potential $(V)$ across $2 \mu F$ is
$V=\frac{Q}{C}=\frac{1200}{2}=600\, V$
$\therefore $ Potential on internal plates
$=1000-600=400\, V$