Answer to Question #133966 in Mechanics | Relativity for sridhar

Question

Consider a parallel plate capacitor which is maintained at potential 200V .
The separation distance between the plates of capacitor and the area of plates are 1mm and 20cm^(2) respectively.Calculate displacement current in 1microsecond .
ANS: 3.5mA

Answer 1

Between the plates of the capacitor, there exists a potential difference, V of 200v. The plates are 1mm apart, and the area of the plates is $20cm^{2}$

Now, V= 20v, distance, d = 1mm = $1\times10^{-3}m$

A = 20cm$^{2} = 20 \times10^{-4}m^{2}$ , and time, t in micro-seconds $\mu s = 10^{-6} s$

A change in electric flux and electric field produces displacement current given by the formula

$I_{d} = \frac{\epsilon_{o} d \Phi_{B}}{dt}$ where $\Phi_{B}$ is electric flux, and $I_{d}$ is the displacement current.

$I_{d} =\frac{ \epsilon_{o} EA}{t}$ where the value of $\epsilon_{o} = 8.85\times10^{-12}$

Electric field, E is given by E = $\frac{V}{d}$

Thus $I= \frac{VI_{d}}{d} = \frac{\epsilon_{o} VA}{td} = \frac{8.85\times10^{-12}\times200\times20\times10^{-4}}{10^{-6}\times1\times10^{-3}}$

$I = 35400 \times 10^{-7} = 3.5 mA$