Published by:
CGP EDU Academic Team
Published on: September 12, 2026
Two parallel plate air capacitors each of capacitance C were connected in series to a battery with e.m.f. ε . Then one of the capacitors was filled up with uniform dielectric with relative permittivity k. How many times did the electric field strength in that capacitor decrease? What amount of charge flows through the battery?
Text Solution
Verified by ExpertsThe correct answer is:
1
(1 + k), Δ q =
Cε 
Sol.

V 1 =
= 
⇒ E 1 =
= 
⇒ E =
= 
E 1 =
E
⇒
(1 + k) time decrease
⇒ q = Q 1 – Q =
= 
⇒ Δ q =
Cε
.
Prepare Smarter with CGP Edu
Get practice questions, solutions, and test series in one place.
Write a Review
Share your experience with this question and solution.
Commentary
Send your comment, doubt, correction, or feedback to admin.
Similar Questions
Explore conceptually related problems
The capacity of a parallel plate condenser is . When a glass plate is placed between the plates of…
A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then sli…
The energy of a charged capacitor is given by the expression ( $\alpha$ = charge on the conductor a…
The capacity of a condenser is $4 \times 10^{-6}$ farad and its potential is 100 volts . The energy…
The insulated spheres of radii $R_1$ and $R_2$ having charges $Q_1$ and $Q_2$ respectively are conn…
In a charged capacitor, the energy resides