The magnitude of the electric field between the two circular parallel plates in
ID: 2211280 • Letter: T
Question
The magnitude of the electric field between the two circular parallel plates in the figure is E = (4.310^5)-(4.010^4*t),
with E in volts per meter and t in seconds. At t = 0, the field is upward as shown. The plate area is 8.010-2m2. For positive times, what is the magnitude of the displacement current?
The magnitude of the electric field between the two circular parallel plates in the figure is E = (4.3½10^5)-(4.0½10^4*t), with E in volts per meter and t in seconds. At t = 0, the field is upward as shown. The plate area is 8.0½10-2m2. For positive times, what is the magnitude of the displacement current?Explanation / Answer
I got id=(jd)A id=(5.94989E-10)*(1.5E-6)=8.92483E-16 And generalized version of Amepre's law states Line integral of B*dl=mu-0(ic+id) ic=(dE/dt)*(A)*(e-0)=8.92483E-16 So the Bc part is Bc=(ic/(2*pi*.088m)) = 2.0283E-21
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.