A source of potential difference V is connected to the combination of two identical capacitors as shown in…

A source of potential difference V is connected to the combination of two identical capacitors as shown in the figure. When key K is closed, the total energy stored across the combination is E1. Now key K is opened and dielectric of dielectric constant 5 is introduced between the plates of the capacitors. The total energy stored across the combination is now E2. The ratio E1E2 will be

  1. 110
  2. 25
  3. 513
  4. 526

Solution

(i) When the switch is closed Ceq=2C

Charge on each capacitor will be q=CV.

Energy E1=12CeqV2 =122C×V2

E1=CV2

(ii) When the switch is opened charge on the right capacitor remain CV while the potential on the left capacitor remains the same.

Dielectric k=5

C'=kC   C'=5C

Now to calculate E2,

E2=12C'V2+q22C'

E2=125CV2+CV225C

E2=5CV22+CV210

E2=13CV25

E1E2=513

Asked in: JEE Main 2022 (26 Jul Shift 2)

Practice more Electrostatics questions on Aicharya