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-established.

Question

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 <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qmg3b71fqyl7vfyo.jpg" /> 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) C V2(K-1) (B) (1/2) CV2(K-1)  (C) C V2(K-1)2 (D) (1/2) C V2(K2-1)

Tardigrade Admin · Accepted Answer

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

U_{i} = \frac{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} = \frac{1}{2} C V^{2}

Heat

= \frac{1}{2} C (K - 1) V^{2}

$C V^{2} (K - 1)$

$\frac{1}{2} C V^{2} (K - 1)$

$C V^{2} (K - 1)^{2}$

$\frac{1}{2} C V^{2} (K^{2} - 1)$

After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is
$U_{i} = \frac{1}{2} K C V^{2}$
After the removal of the lab and steady-state is reached, the energy stored in the capacitor is
$U_{f} = \frac{1}{2} C V^{2}$
Heat $= \frac{1}{2} C (K - 1) V^{2}$

CV2(K−1)

21​CV2(K−1)

CV2(K−1)2

21​CV2(K2−1)

After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is Ui​=21​KCV2 After the removal of the lab and steady-state is reached, the energy stored in the capacitor is Uf​=21​CV2 Heat =21​C(K−1)V2

$C V^{2} (K - 1)$

$\frac{1}{2} C V^{2} (K - 1)$

$C V^{2} (K - 1)^{2}$

$\frac{1}{2} C V^{2} (K^{2} - 1)$

After the insertion of the slab and steady-state is reached, the energy stored in the capacitor is
$U_{i} = \frac{1}{2} K C V^{2}$
After the removal of the lab and steady-state is reached, the energy stored in the capacitor is
$U_{f} = \frac{1}{2} C V^{2}$
Heat $= \frac{1}{2} C (K - 1) V^{2}$