Answer:
![1.7\cdot 10^{15} V m^{-1} s^{-1}](https://tex.z-dn.net/?f=1.7%5Ccdot%2010%5E%7B15%7D%20V%20m%5E%7B-1%7D%20s%5E%7B-1%7D)
Explanation:
The electric field between the plates of a parallel-plate capacitor is given by
(1)
where
V is the potential difference across the capacitor
d is the separation between the plates
The potential difference can be written as
![V=\frac{Q}{C}](https://tex.z-dn.net/?f=V%3D%5Cfrac%7BQ%7D%7BC%7D)
where
Q is the charge stored on the plates of the capacitor
C is the capacitance
So eq(1) becomes
(2)
Also, the capacitance of a parallel-plate capacitor is
![C=\frac{\epsilon_0 A}{d}](https://tex.z-dn.net/?f=C%3D%5Cfrac%7B%5Cepsilon_0%20A%7D%7Bd%7D)
where
is the vacuum permittivity
A is the area of the plates
Substituting into (2) we get
(3)
Here we want to find the rate of change of the electric field inside the capacitor, so
![\frac{dE}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdE%7D%7Bdt%7D)
If we calculate the derivative of expression (3), we get
![\frac{dE}{dt}=\frac{1}{\epsilon_0 A}\frac{dQ}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7BdE%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7B%5Cepsilon_0%20A%7D%5Cfrac%7BdQ%7D%7Bdt%7D)
However,
corresponds to the definition of current,
![I=\frac{dQ}{dt}](https://tex.z-dn.net/?f=I%3D%5Cfrac%7BdQ%7D%7Bdt%7D)
So we have
![\frac{dE}{dt}=\frac{I}{\epsilon_0 A}](https://tex.z-dn.net/?f=%5Cfrac%7BdE%7D%7Bdt%7D%3D%5Cfrac%7BI%7D%7B%5Cepsilon_0%20A%7D)
In this problem we have
I = 3.9 A is the current
is the area of the plates
Substituting,
![\frac{dE}{dt}=\frac{3.9}{(8.85\cdot 10^{-12})(2.56\cdot 10^{-4})}=1.7\cdot 10^{15} V m^{-1} s^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7BdE%7D%7Bdt%7D%3D%5Cfrac%7B3.9%7D%7B%288.85%5Ccdot%2010%5E%7B-12%7D%29%282.56%5Ccdot%2010%5E%7B-4%7D%29%7D%3D1.7%5Ccdot%2010%5E%7B15%7D%20V%20m%5E%7B-1%7D%20s%5E%7B-1%7D)