Question

A parallel-plate capacitor has plate are 20 cm^{2, plate separation 1.0 mm and a dielectric slab of dielectric constant 5.0 filling up the space between the plates. This capacitor is joined to a battery of emf 6.0 V through a 100 k $\Omega$ resistor. Find the energy of the capacitor 8.9 $\mu s$after the connections are made.

Answer 1

$A=20c{m}^2,d=1mm,K=5,$

$t=8.9\times 1^{-6}C,ϵ=6V$

$R=100\times {10}^{-3}\Omega$

$C=\frac{K{ϵ}_{0}A}{d}$

$=\frac{5\times 8.85\times {10}^{-12}\times 20\times {10}^{-4}}{1\times {10}^{-3}}$

$=88.5\times {10}^{-12}$

$Q=EC(1-e^{\frac{-t}{rc}})$

$=6\times 88.5\times {10}^{-12}(1-e)$

$=530.97\times {10}^{-12}C$

$\text{Energy}=\frac{1}{2}\times \frac{500\times 530}{88\times {10}^{12}}$

$=\frac{500\times 530.97}{88\times 2}\times {10}^{-12}$