Question

A parallel-plate capacitor has plate area 40 $c{m}^2$, and separation between the plates 0.10 mm is connected to a battery of emf 2.0 V through a 16 $\Omega$ resistor. Find the electric field in the capacitor 10 ns after the connections are made.

Answer 1

$A=40c{m}^2=40\times {10}^{-4}{m}^2$

$D=0.1mm=1\times {10}^{-4}m,$

$R=16\Omega ,E_{mf}=2V$

$C=\frac{{ϵ}_{0}A}{d}$

$=\frac{8.85\times {10}^{-12}\times 40\times {10}^{-4}}{1\times {10}^{-4}}$

$=35.4\times {10}^{-11}F$

$\text{Now,}E=\frac{Q}{A{E}_0}(1-e^{\frac{-t}{rc}})$

$=\frac{35.4\times {10}^{-11}\times 2}{8.85\times {10}^{-12}\times 40\times {10}^{-4}}(1-e^{-1.76})$

$=1.655\times {10}^{-4}=1.7\times {10}^{-4}V/m$