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

`(dE)/(dt)=2.07*10^(13)(V/m)/(s)`

Step-by-step 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=(\epsilon_0 \mu_o)/(2\pi r)(d\Phi)/(dt)(2)`

Replacing (1) in (2) and solving for
`(dE)/(dt)`:

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