this is a top down view of a circular table top of radius R=1m. An electric fiel
ID: 2077854 • Letter: T
Question
this is a top down view of a circular table top of radius R=1m. An electric field points upward through the table-top perpendicular to the surface ( out of the page in the drawing) the field is strongest near the center and weakest near the edge: E(r)= 2 v/m^2 (R-r)
where r is the distance from the ceenter of the table. Suppose this field is steadily reduced to zero over a three second interval
A) Show that the initial flux through the table top is e=2/3 pi R^3, where the “2/3” has units of volts/meter squared.
B)During the three-second interval, what is the magnitude of the magnetic field at point P
C)During the three-second interval, what is the displacement current through the tabletop?
Explanation / Answer
R = 1m
E = [2(R-r)] V/m^2 z
dE/dt = 2(R-r)/3
a)Flux F = integral(E.dA) = integral(2(R-r)* 2*pi*r dr) from 0 to R
F = 4*pi(R*r^2/2 - r^3/3) from 0 to R
F = 4*pi*(R^3)/6 = [(2/3) * pi * R^3] V-m
b) integral(B.dl) = (mu)*(epsilon)* dF/dt
Since E-field is radially symmetric
=> B * 2*pi*R = 2 * pi/(3*3*9 * 10^16)
=> B = 1.12 * 10^-18 T
c) Displacement current Id = epsilon * dF/dt = 8.85 ^ 10^-12 * 2*pi/(3*3) = 0.616 ^ 10^-11 A
d) B = mu * I/(2*pi*R)
=> I = 1.23 * 10^-18 * 2*pi/ (4*pi * 10^-7) = 0.615 * 10^-11 A
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