An electric field has strength E = (65t - 22t²)V/m in a region of space.What is the displacement current through a circular area of diameter 15 cm in the x-y plane in this region at t = 5.0s?

Question

Quizplus · Accepted Answer

The displacement current through a circular area is given by:

$I_D = \epsilon_0\frac{d}{dt}\int\int_S E\cdot dA$

We are given that the electric field is:

$E = (65t - 22t^2)V/m$

To calculate the displacement current through a circular area of diameter 15 cm in the x-y plane, we need to calculate the integral of $E\cdot dA$ over the circular area. Since the electric field is in the z-direction, we can take $dA$ to be a vector in the z-direction. Therefore, $E\cdot dA = (65t - 22t^2)dA_z$.

The circular area has a diameter of 15 cm, which means it has a radius of 7.5 cm. We can write $dA_z$ as $rd	heta dz$, where $r$ is the radius of the circular area, $	heta$ is the angle between the normal to the circular area and the z-axis, and $dz$ is a small element of the z-direction. Integrating over the circular area gives:

$I_D = \epsilon_0\frac{d}{dt}\int\limits_{0}^{2\pi}\int\limits_{0}^{7.5}\left[(65t - 22t^2)right]r\,d	heta dz$

$I_D = \epsilon_0\frac{d}{dt}\left[\pi\int\limits_{0}^{7.5}\left[(65t - 22t^2)r^2ight]dzight]$

$I_D = \epsilon_0\frac{d}{dt}\left[\pi\left(\frac{65}{3}t - \frac{11}{3}t^3ight)7.5^3ight]$

$I_D = \epsilon_0\pi\left(\frac{65}{3} - 11t^2ight)7.5^2\frac{d}{dt}(t)$

$I_D = \frac{1}{4}\pi\epsilon_0\left(975 - 165t^2ight)$

At $t = 5.0s$, the displacement current is:

$I_D = \frac{1}{4}\pi\epsilon_0\left(975 - 165(5.0)^2ight) = \frac{1}{4}\pi(8.85	imes 10^{-12})(-3875) \approx \boxed{24\,	ext{pA}}$

An Electric Field Has Strength E = (65t - 22t2)V/m

An Electric Field Has Strength E = (65t - 22t²)V/m