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NTA Abhyas JEE Main2020PhysicsCurrent ElectricityPractice

An air-filled parallel plate capacitor has capacitance C . The capacitor is connected through a resistor to a voltage source providing a constant potential difference V A dielectric plate with a dielectric constant K is inserted into the capacitor, filling it completely. After the equilibrium is established plate is quickly removed. Find the amount of heat generated in the resistor by the time, the equilibrium is re-

Options

  1. AC V 2   K -   1
  2. B1 2 C V 2 K - 1
  3. CC V 2   K -   1 2
  4. D1 2 C V 2 K 2 -   1

Correct answer

B. 1 2 C V 2 K - 1

Step-by-step solution

After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is U i = 1 2 K C V 2 After the removal of the lab and steady-state is reached, the energy stored in the capacitor is U f = 1 2 C V 2 H e a t = 1 2 C K - 1 V 2

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