a flat circle of radius 18 cm is placed in a uniform electric field of magnitude 5.8 x 10^2 N/C. What is the electric flux through the circle when its face is (a) perpendicular to the field lines, (b) at 45 degree angle to the field lines, and (c) parallel to the field lines?
How do you figure this out?
2007-01-18 05:56:16 · 4 answers · asked by Anonymous in Science & Mathematics ➔ Physics
The area integral of the electric field over the area of this circle of radius 18 cm is equal to the net charge enclosed by the surface divided by the permittivity. I'll leave the arithmetic and the closed integration to you. This would be for case (a). At a 45 deg angle (a)would be multiplied by sin 45 and for (c) the sin 0 is zero. Good luck.
2007-01-18 06:11:06 · answer #1 · answered by 1ofSelby's 6 · 0⤊ 0⤋
I don't remember the details of the formula to calculate flux, but I recall that the Flux is a measure of the "amount" of electric filed that can pass through a certain area. If it is perpendicular to the direction of the Field lines, this will maximize Flux.
This website should be very helpful though.
2007-01-18 06:00:56 · answer #2 · answered by Michael Dino C 4 · 0⤊ 0⤋
r = 0.18 m
E = 5.8 * 10^2 N/C
Flux
a theta = 0,
flux = EA = 5.8*10^2 * 0.18 = 1.044 * 10^2
b theta = 45º, cos theta = 1/root 2
flux = EA cos theta = 0.735*10^2
c flux = 0
2007-01-18 06:13:16 · answer #3 · answered by Aditya Joshi 2 · 0⤊ 0⤋
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2016-11-25 01:38:51 · answer #4 · answered by Anonymous · 0⤊ 0⤋