Question

Two capacitors of capacitance 2μF and 3μF 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

Answer 1

Equivalent capacitance

\(= \frac{2 \times 3}{2+3} = \frac{6}{5} \mu\)

Total charge, Q = CV = \(\frac{6}{5} \times 1000\)

= 1200 μC

Potential (V) across 2μF is

\(V = \frac{Q}{C} = \frac{1200}{2} = 600 V\)

Potential on internal plates

= 1000 - 600 = 400 V