Question

The induced magnetic field at radial distance 4.0 mm from the central axis of a circular parallel-plate capacitor is 1.8 ✕ 10−7 T. The plates have radius 2.5 mm. At what rate dE with arrow/dt is the electric field between the plates changing?

Answer 1

Answer:

$\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$

Explanation:

According to Gauss's law, the electric flux through the circular plates is defined as the electric field multiplied by its area:

$\Phi=EA=E(\pi R^2)(1)$

The magnetic field around the varying electric field of the circular plates is given by:

$B=\frac{\epsilon_0 \mu_o}{2\pi r}\frac{d\Phi}{dt}(2)$

Replacing (1) in (2) and solving for $\frac{dE}{dt}$:

$B=\frac{\epsilon_0 \mu_o\pi R^2}{2\pi r}\frac{dE}{dt}\\\frac{dE}{dt}=\frac{2rB}{\epsilon_0 \mu_o R^2}\\\frac{dE}{dt}=\frac{2(4*10^{-3}m)(1.8*10^{-7}T)}{(8.85*10^{-12}\frac{C^2}{N\cdot m^2})(4\pi *10{-7}\frac{Tm}{A})(2.5*10^{-3}m)^2}\\\\\frac{dE}{dt}=2.07*10^{13}\frac{V/m}{s}$