Q. Condenser $A$ has a capacity of $15\mu F$ when it is filled with a medium of dielectric constant 15. Another condenser $B$ has a capacity of $1\pi F$ with air between the plates. Both are charged separately by a battery of $100\ V.$ After charging, both are connected in parallel without the battery and the dielectric medium being removed. The common potential now is
A) $400\ V$
B) $800\ V$
C) $1200\ V$
D) $1600\ V$
Explanation
(b) Charge of capacitor $A$ is given by
$Q_{1} = 15 \times 10^{- 6} \times 100 = 15 \times 10^{- 4}C$
Charge on capacitor $B$ is given by
$Q_{2} = 1 \times 10^{- 6} \times 100 = 10^{- 4}C$
Capacity of capacitor $A$ after removing dielectric
$= \frac{15 \times 10^{- 6}}{15} = 1\mu F$
Now when both capacitors are connected in parallel their equivalent capacitance will be $C_{eq} = 1 + 1 = 2\mu F$
So common potential $= \frac{\left( 15 \times 10^{- 4} \right) + (1 \times 10^{- 4})}{2 \times 10^{- 6}} = 800V$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.