Q36 / 143 · NTA Abhyas (JEE Main) 2020 · Physics · Current Electricity

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

Figure

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-established.

  • A
    CV2 K- 1
  • Correct answer
    B
    12CV2 K- 1
  • C
    CV2 K- 12
  • D
    12CV2K2- 1

Solution

After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is

Ui=12KCV2

After the removal of the lab and steady-state is reached, the energy stored in the capacitor is

Uf=12CV2

Heat=12CK- 1V2

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

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