Q. Two capacitor 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) $700\ Volt$
B) $200\ Volt$
C) $600\ Volt$
D) $400\ Volt$
Explanation
(d) Equivalent capacitance $= \frac{2 \times 3}{2 + 3} = \frac{6}{5}\mu F$
Total charge by $Q = CV = \frac{6}{5} \times 1000 = 1200\mu C$
Potential $(V)$ across $2\mu F$ is $V = \frac{Q}{C} = \frac{1200}{2} = 600volt$
$\therefore$ Potential on internal plates $= 1000 - 600 = 400\ V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.