Answer:
The magnitude of rate of change of electric field is [tex]49.95\ V/m{\cdot} s[/tex].
Explanation:
Given that,
Radius of the cylindrical region contains a uniform electric field along the cylinder axis, r = 1.2 m
Total displacement current through a cross section of the region, [tex]I=2\times 10^{-9}\ A[/tex]
We need to find the rate of change of electric field. Its is given by the formula as follows :
[tex]\dfrac{dE}{dt}=\dfrac{I}{A\epsilon_o}\\\\\dfrac{dE}{dt}=\dfrac{2\times 10^{-9}}{\pi (1.2)^2\times 8.85\times 10^{-12}}\\\\\dfrac{dE}{dt}=49.95\ V/m{\cdot} s[/tex]
So, the magnitude of rate of change of electric field is [tex]49.95\ V/m{\cdot} s[/tex].
