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QJEEM23Pha1bc: JEE Main 2023 Physics - Class 12 JEEM23Pha1bc Solution: Two parallel plate capacitors and each having capacitance of…

Two parallel plate capacitors C1C_{1} and C2C_{2} each having capacitance of 10μF10 \mu \mathrm{F} are individually charged by a 100 V D.C. source. Capacitor C1C_{1} is kept connected to the source and a dielectric slab is inserted between it plates. Capacitor C2\mathrm{C}_{2} is disconnected from the source and then a dielectric slab is inserted in it. Afterwards the capacitor C1C_{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=10 )

Step-by-Step Verified Solution

Charge on C1=KCV\mathrm{C}_{1}=\mathrm{KCV}

And charge on C2=CV\mathrm{C}_{2}=\mathrm{CV}

When they are connected in parallel charge will be equally divided so charge on one capacitor is

q=K+12CVq=\frac{K+1}{2} \mathrm{CV}

So, V=qKC=K+12K=55 VV=\frac{q}{K C}=\frac{K+1}{2 K}=55 \mathrm{~V}

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