Prove that the field strength inside the Faraday cylinder far from its ends is zero. ??

Expert Solution

Let us assume that there exists some field inside the cylinder. According to symmetry that field must be radially outwards (or inwards). Now we apply the Gauss' law to the Gaussian surface shown in the figure.

as the charge inside the cylinder is zero.

Also note that at every point on the flat two surfaces on the given Gaussian surface, field is parallel to area and hence perpendicular to area vector. So flux through those two surfaces is zero. Also for any small area on the curved part of the Gaussian surface under consideration is perpendicular to the field and hence area vector is parallel to the field. So angle between area vector and field is 0⁰. Moreover at every point on the considered Gaussian surface, the magnitude of the field is same (again by symmetry). Hence we get:

But we already know that this integral must be zero (from 1).

Hence E = 0.

genius_generous answered 8 months ago

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