Electrostatics Question 569

Question: Two identical parallel plate capacitors are connected in series to a battery of 100$ V $ . A dielectric slab of dielectric constant 4.0 is inserted between the plates of second capacitor. The potential difference across the capacitors will now be respectively [MP PMT 1992]

Options:

A) 50 V, 50 V

B) 80 V, 20 V

C) 20 V, 80 V

D) 75 V, 25 V

Show Answer

Answer:

Correct Answer: B

Solution:

$ C _{eq}=\frac{C\times 4C}{(C+4C)}=\frac{4C}{5} $

$ Q=C _{eq}.V=\frac{4C}{5}\times 100=80C $

Hence $ V _{1}=\frac{Q}{C _{1}}=\frac{80C}{C _{1}}=80V $ and $ V _{2}=\frac{80C}{4C}=20V $



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