Would the magnitude of the electric field at a point away from the center of a sphere depend on the radius of the sphere?

Submitted by cozymikey t3_11b52ll in askscience

Suppose you have two conducting metal spheres with the same negative charge, however, one has a radius twice that of the other's. If I wanted to determine the electric field at a point away from the center of each sphere, which sphere would should have a larger calculated electric field?

When using the electric field formula E=kq/r^2, it seems that the radius of the sphere shouldn't have an effect on the calculated electric field. Is there a chance that the r in the equation in this case would actually represent the distance between the surface of the sphere and the point, rather than the center?

Any help would be greatly appreciated, thanks :)

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Movpasd t1_j9ysumc wrote

> Is there a chance that the r in the equation in this case would actually represent the distance between the surface of the sphere and the point, rather than the center?

The charges for a conducting sphere distribute themselves uniformly on the surface of the sphere. Each little element of charge on the surface contributes an infinitesimal amount of the final electric field. To calculate the final field, you need to (vector) integrate the contributions from all these elements. So you can't just use kq/r^2 but with r the "altitude" of the test charge.

If you do this calculation, you'll find that it actually can apply E = kq/r^(2) with r the distance to the centre of the sphere -- the uneven contributions cancel out. From the outside, a spherical shell of uniform charge looks exactly like a point charge at its centre.

This is actually true for any company spherical charge distribution, and you can prove it very elegantly using Gauss's law.

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kilotesla t1_ja0lrto wrote

Replying to Movpasd (#1,996,487)

This is a good answer. Like the other answers, it assumes that the sphere of metal is far away from other charged or polarizable objects. If the two spheres being discussed were close to each other, the charge would no longer be uniformly distributed on either of them and the analysis isn't as simple.

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