At what distance along the central axis of a uniformly charged plastic disk of radius R = 1.37 m is the magnitude of the electric field equal to 1/5 times the magnitude of the field at the center of the surface of the disk?

Question

roadjester · Answer

do you have the correct answer?

austinl · Answer

No, no I do not.

anonymous · Answer

When you find the formula for the exact answer, 
you could  compare it against
the  upper limit of E = surface charge density 
and lower limit of E = K q/r^2 as though the disk were a point charge.

roadjester · Answer

$$\huge E=2\pi k_e\sigma$$
where k_e is coulomb's constant and sigma is the surface charge density

anonymous · Answer

I think the point is that the field falls off with distance. The correct formula should have distance along the axis as part of it.

roadjester · Answer

well ok, I just assumed that the distance was 0

roadjester · Answer

and reduced the equation

roadjester · Answer

$$\huge E_x=2 \pi k_e \ \sigma {[1-{x\over (R^2 +x^2)^{1\over 2}}]}$$

roadjester · Answer

this disk is on the yz-plane
if you have a point charge on the x-axis then x represents the distance the charge is away from the disk
R is the radius of the disk

roadjester · Answer

did you actually need help deriving the equation for the E-field?