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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 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)

Option: 1

55


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

1.


2.


3.

By charge conservation

\mathrm{Q}_{1}=\mathrm{Q}_{2}

\mathrm{KC}_{1} \mathrm{~V}+\mathrm{C}_{2} \mathrm{~V}=\left(\mathrm{KC}_{1}+\mathrm{KC}_{2}\right) \mathrm{V}_{\text {common }}

\mathrm{V}_{\text {common }}=\frac{(\mathrm{K}+1) \mathrm{CV}}{2 \mathrm{KC}}=\frac{\mathrm{K}+1}{2 \mathrm{~K}} \mathrm{~V}

\mathrm{V}_{\text {common }}=\frac{11}{20} \times 100=55 \mathrm{~V}
 

Posted by

Riya

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