(a) [tex]1.49\cdot 10^{12} Vm^{-1}s^{-1}[/tex]
We have:
r = 11.0 cm = 0.11 m is the radius of the plates, so their area is
[tex]A=\pi r^2 = \pi (0.11 m)^2=0.038 m^2[/tex]
The plate separation is
d = 4.00 mm = 0.004 m
The potential difference across the plates of the capacitor is
[tex]V=\frac{Q}{C}[/tex] (1)
where Q is the charge and C the capacitance; and since
[tex]V=Ed[/tex]
(E= electric field), we can rewrite (1) as
[tex]E=\frac{Q}{dC}=\frac{Q}{d(\frac{\epsilon_0 A}{d})}=\frac{Q}{\epsilon_0 A}[/tex]
So, the rate variation of the electric field is given by
[tex]\frac{dE}{dt}=\frac{1}{\epsilon_0 A}\frac{dQ}{dt}=\frac{I}{\epsilon_0 A}[/tex]
Since [tex]I=\frac{dQ}{dt}[/tex] (the current is the rate of change of the charge).
Substituting all the numbers, we find
[tex]\frac{dE}{dt}=\frac{0.500 A}{(8.85\cdot 10^{-12} F/m)(0.038 m^2)}=1.49\cdot 10^{12} Vm^{-1}s^{-1}[/tex]
(b) [tex]4.1\cdot 10^{-7} T[/tex]
The magnetic field between the plates of the capacitor can be found by using Maxwell's equations, and it is given by
[tex]B=\frac{\mu_0 d I}{2 A}[/tex]
where d is the radial distance from the centre.
In this case, we have
d = 5.00 cm = 0.05 m
Therefore, the magnetic field is
[tex]B=\frac{(12.56\cdot 10^{-7} H/m)(0.05 m)(0.500 A)}{2(0.038 m^2)}=4.1\cdot 10^{-7} T[/tex]
