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 $
BETA
