Question

How can I prove that the electric field inside an irregular cavity (not spherical) with no charge, of a charged conductor is zero?

Answer 1

Since you have a conductor, all the charge must lie on the outer edge of the object. So in the cavity there is no charge, because all the charge is on the conductor. Make any closed Gaussian surface within the cavity (the shape does not matter as long as it is close), and you will enclose no net charge. Therefore by Gauss's law the electric field must be zero within the cavity.

Answer 2

You must know the definition of Gaussian surface:
around a charge distribution, is a closed surface, such that electric field intensity at all the points on the surface is same and the electric flux through the surface is along the normal to the surface.

If i draw an irregular surface it's not a Gauss surface.

Answer 3

electric field intensity thus on the surface will be unequal due to unequal values of r^2.

Answer 4

If there is no charge inside the closed surface, then flux is zero and fields of outside charge or external field will not be given by any flux. Hence ,electric field must be zero within the cavity .

Answer 5

@krikrista, everyone can say that; but I need to prove that- not just make some petty assumption.

Answer 6

Hi

Answer 7

Well, for the proof I need to ponder upon that but for the time-being, when any conductor is placed inside a region of uniform electric field, the free electrons get drifted in opposite direction resulting into+ve and -ve charge densities on opposite faces of metal. This charge density results into an internal electric field in opposite direction to that of applied field and in equillibrium the net field becomes zero.

Answer 8

Yeah, that I know; but the hint provided with the question says that the previous theory won't be applicable in this case since assumptions cannot be made for irregular surface without concrete proof.

And I figured out that the field that comes out from the inner surface of the spherical cavity cancels out itself as all the fields tend to converge on the same point. But in case of irregular surface, the tangents drawn won't converge at the same point- so there's a possible chance of a prevailing field; until proved otherwise.

You can see the point I'm trying to make from these drawings
:

But,