Two capacitors of capacitance 2 μ F and 3 μ 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

Question

Two capacitors of capacitance 2 μ F and 3 μ 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

Tardigrade Admin · Accepted Answer

Equivalent capacitance =(2 × 3/2+3)=(6/5) μ Total charge, Q=C V=(6/5) × 1000 =1200 μ C Potential (V) across 2 μ F is V=(Q/C)=(1200/2)=600 V ∴ Potential on internal plates =1000-600=400 V

Q. Two capacitors of capacitance $2 μ F$ and $3 μ F$ 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

700 V

200 V

600 V

400 V

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

Q. 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