Answer:
a
[tex]\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s[/tex]
b
[tex]I = 0.106 \ A[/tex]
Explanation:
From the question we are told that
The current is [tex]I = 0.106 \ A[/tex]
The length of one side of the square [tex]a = 4.60 \ cm = 0.046 \ m[/tex]
The separation between the plate is [tex]d = 4.0 mm = 0.004 \ m[/tex]
Generally electric flux is mathematically represented as
[tex]\phi_E = \frac{Q}{\epsilon_o}[/tex]
differentiating both sides with respect to t is
[tex]\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} * \frac{d Q}{ dt}[/tex]
=> [tex]\frac{d \phi_{E}}{dt} = \frac{1}{\epsilon_o} *I[/tex]
Here [tex]\epsilon_o[/tex] is the permitivity of free space with value
[tex]\epsilon _o = 8.85*10^{-12} C/(V \cdot m)[/tex]
=> [tex]\frac{d \phi_{E}}{dt} = \frac{0.106}{8.85*10^{-12}}[/tex]
=> [tex]\frac{d \phi_{E}}{dt} =1.1977 *10^{10} \ V\cdot m/s[/tex]
Generally the displacement current between the plates in A
[tex]I = 8.85*10^{-12} * 1.1977 *10^{10}[/tex]
=> [tex]I = 0.106 \ A[/tex]
