# What is the electric field and potential outside a spherical capacitor?

Two concentric spheres form a spherical capacitor with the same charges (but opposite signal). I know, by Gauss's law, that the electric field must be zero (actually, the flux must be zero, but I can't see how the flux can be zero and the electric field is not zero). But why there is no net field outside the spherical capacitor if the negative charges in the (for example) external sphere create a electric field in every direction - including de direction pointing to outside the sphere? (in this case, there is no other electric that could cancel it).

So I think it's supposed to be like in this picture - and the external electric is falling off with the distance, as any electric field, and the electric field of the charge inside makes it fall off faster - though it is never zero.

If it is so, why does Gauss's law says it is zero? If it's not, why? (in this latter case, how can we explain it NOT using Gauss's law, using only the charges and electric fields?) I suppose the same holds for cylindrical capacitors

• The total charge is zero, so the electric field outside the shell is zero. If the center and the shell don't carry the opposite charge, then the field outside is not zero. – CuriousOne Aug 1 '16 at 23:27
• "But why there is no net field outside the spherical capacitor if the negative charges in the (for example) external sphere create a electric field in every direction - including de direction pointing to outside the sphere?" - Have you considered using superposition? – Alfred Centauri Aug 1 '16 at 23:31
• Alfred Centauri, yes I did and since the points outside the external sphere are closer to the the external sphere than the inside sphere, the "negative electric fiel" (electric field of the external sphere) is stronger than the "positive field" in the points outside the sphere. So the fields have the opposite directions and at first they could cancel each other but they don't because their magnitudes are different. – João Paulo Aug 1 '16 at 23:46
• CuriousOne, this is not true in general. The net charge in a dipole is zero though its electric field is not zero. Why would it be so with the shells? – João Paulo Aug 1 '16 at 23:49

Alfred Centauri, yes I did [consider using superposition] and since the points outside the external sphere are closer to the the external sphere than the inside sphere, the "negative electric fiel" (electric field of the external sphere) is stronger than the "positive field" in the points outside the sphere. So the fields have the opposite directions and at first they could cancel each other but they don't because their magnitudes are different.

In the exterior region of an uniformly charged sphere, centered on the origin and with total charge Q, the electric field is identical to that of a point charge Q located at the origin.

Thus, by superposition, the electric field in the region outside of the concentric, uniformly charged spheres is simply the electric field due a point charge at the origin with charge equal to the sum of the total charge on each sphere.

In the case that the spheres have equal and opposite total charge, the sum is zero and, thus, the electric field in the exterior region is zero.

• Thank you, I understood it. I have thought about this explanation previously, but though this make sense and Gauss's law surely give me the correct answer (the electric field outside is 0 ), I wasn't sure because I couldn't find the failure in my reasonig that the electric field outside would not be zero. (But now I do). Thanks. – João Paulo Aug 2 '16 at 23:54

So I think it's supposed to be like in this picture - and the external electric is falling off with the distance, as any electric field, and the electric field of the charge inside makes it fall off faster - though it is never zero.

The electric field for a sphere is: $$E=\frac{kQ}{R}$$. The inner sphere has a smaller $$R$$ than the larger sphere, but $$E$$ is larger for the inner sphere. The inner sphere does cancel the weaker $$E$$ from the larger sphere outside the larger sphere.

• Please use Mathjax when typing mathematical expressions. – Andrew Apr 28 at 4:02