Question 390: Electrostats & Capacitance

Question

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 $=10$ )

Accepted Answer

Charge on $\mathrm{C}_{1}=\mathrm{KCV}$
And charge on $\mathrm{C}_{2}=\mathrm{CV}$
When they are connected in parallel charge will be equally divided so charge on one capacitor is
$q=\frac{K+1}{2} \mathrm{CV}$
So, $V=\frac{q}{K C}=\frac{K+1}{2 K}=55 \mathrm{~V}$

Electrostats & Capacitance

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