Q75 / 90 · JEE Main 2022 (26 Jul Shift 2) · Physics

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

Figure

  • A
    110
  • B
    25
  • Correct answer
    C
    513
  • D
    526

Solution

Figure

(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

Hence, the correct answer is option: 'C'.

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