Q. A potential difference of $300\ volts$ is applied to a combination of $2.0\mu F$ and $8.0\mu F$ capacitors connected in series. The charge on the $2.0\mu F$ capacitor is
A) $2.4 \times 10^{- 4}C$
B) $4.8 \times 10^{- 4}C$
C) $7.2 \times 10^{- 4}C$
D) $9.6 \times 10^{- 4}C$
Explanation
(b) In series combination charge $Q$ is same. So charge on $2\mu F$ capacitor is
$Q = C_{eq}V = \left( \frac{2 \times 8}{2 + 8} \right) \times 300 \times 10^{- 6} = 4.8 \times 10^{- 4}C$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.