Electric field question?

Question

A solid metal sphere with radius 0.410 m carries a net charge of 0.280 nC.

Find the magnitude of the electric field at a point 0.111 m outside the surface of the sphere.

The internet says my answer is off by an additive constant but I cant figgure out what?

My solution (.280X10^-9) / [ (4)(Pi)(E0)(.111)^2

Answer 1

What is your solution?

EDIT

I found it. Use Guass' Law to determine the electric field from the sphere. Guass' Law essentially states that:

E (4 pi r^2) = q / e0

where q is the total charge enclosed by the sphere with a radius r

So,

E = q / (e0 4 pi r^2), which is basically the same thing you have with one minor mistake. Using Gauss' Law models the charge distribution as a point charge at the center of the solid metal sphere. Of course, this only works for points outside the sphere as the electric field in a conductor is zero. This changes your value of r.

q = 0.280 nC
r = 0410 m + 0.111 m = 0.521 m

Try this one out and good luck.

Answer 2

The field at r is the same as that for all the surface charge to be located at a point at the center of the sphere.