Question

An air filled parallel plate capacitor has capacitor C. The capacitor is connected through a resistor to a voltage source providing a constant potential difference V. A dielectric plate with 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 reestablished –

• A. CV² (K – 1)
• B. $\frac{1}{2}$CV² (K – 1)²
• C. CV² (K – 1)²
• D. $\frac{1}{2}$CV² (K² – 1)

Answer 1

Answer: (B)

$\frac{1}{2}$CV² (K – 1)²

Answer 2

t = 0 q = KCV

iR + V – $\frac{q}{C}$ = 0

i = – $\frac{dq}{dt}$

–$\frac{dq}{dt}$ × R + V – $\frac{q}{C}$ = 0

–$\frac{dq}{dt}$ × RC = q – CV

$\frac{dq}{q - CV}$ = – $\frac{dt}{RC}$

log (q – CV) = – $\frac{t}{RC}$ + C

t = 0

q = KCV

C = log (KCV – CV)

log $\left\lbrack \frac{q - CV}{KCV - CV} \right\rbrack$ = – $\frac{t}{RC}$

q = (K – 1) CVe^–t/RC + CV

i = – $\frac{dq}{dt}$ ⇒ i = $\frac{+ (K - 1)CVe^{- t/RC}}{RC}$

Heat = $\int_{0}^{\infty}{i^{2}Rdt}$

Heat =

=$\left\lbrack - RCR \times \frac{(K - 1)^{2}V^{2}}{R^{2}}\frac{e^{- \frac{2t}{RC}}}{2} \right\rbrack_{0}^{\infty}$Heat = $\frac{1}{2}$C (K – 1)² V²