JEE Main & Advanced
Physics
Electrostatics & Capacitance
Question Bank
Grouping of Capacitors
question_answer
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]
A) 50 V, 50 V
B) 80 V, 20 V
C) 20 V, 80 V
D) 75 V, 25 V
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 \[{{V}_{2}}=\frac{80C}{4C}=20V\]