JEE Main202226 Jul 2022Evening ShiftPhysicsCapacitanceActual
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 E 1 . 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 E 2 . The ratio E 1 E 2 will be
Options
- A1 10
- B2 5
- C5 13
- D5 26
Correct answer
C. 5 13
Step-by-step solution
(i) When the switch is closed C eq = 2 C Charge on each capacitor will be q = C V . Energy E 1 = 1 2 C eq V 2 = 1 2 2 C × V 2 E 1 = C V 2 (ii) When the switch is opened charge on the right capacitor remain C V while the potential on the left capacitor remains the same. Dielectric k = 5 C ' = k C C ' = 5 C Now to calculate E 2 , E 2 = 1 2 C ' V 2 + q 2 2 C ' ⇒ E 2 = 1 2 5 C V 2 + C V 2 2 5 C ⇒ E 2 = 5 C V 2 2 + C V 2 10 ⇒ E 2 = 13 C V 2 5 ⇒ E 1 E 2 = 5 13