Two parallel plate capacitors $C_{1}$ and $C_{2}$ each having capacitance of $10 \mu \mathrm{F}$ are individually charged by a 100 V D.C. source. Capacitor $C_{1}$ is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor $\mathrm{C}_{2}$ is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor $C_{1}$ is also disconnected from the source and the two capacitors are finally connected in parallel combination. The common potential of the combination will be ________ V.
(Assuming Dielectric constant $=10$ )
Answer (integer)
55
Solution
Charge on $\mathrm{C}_{1}=\mathrm{KCV}$
<br/><br/>And charge on $\mathrm{C}_{2}=\mathrm{CV}$
<br/><br/>When they are connected in parallel charge will be equally divided so charge on one capacitor is
<br/><br/>$q=\frac{K+1}{2} \mathrm{CV}$
<br/><br/>So, $V=\frac{q}{K C}=\frac{K+1}{2 K}=55 \mathrm{~V}$
About this question
Subject: Physics · Chapter: Electrostatics · Topic: Electric Potential and Capacitance
This question is part of PrepWiser's free JEE Main question bank. 132 more solved questions on Electrostatics are available — start with the harder ones if your accuracy is >70%.