A hollow, conducting sphere with an outer radius of 0.260 m and an inner radius

Question

A hollow, conducting sphere with an outer radius of 0.260 m and an inner radius of 0.200 m has a uniform surface charge density of + 6.37 times 10^-6 C/m^2. A charge of -0.580 pC is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? 5.3e-6 C/m^2 (b) Calculate the strength of the electric field just outside the sphere. N/C (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere? N middot m^2/C

Explanation / Answer

Part a

As due to induction an equal and opposite charge (+ 0.580 C) appears on the inner surface of the sphere and hence (- 0.580 C) appears on the outer surface of sphere.

So induced charge density on outer surface is = -0.580x10-6/*(0.260)2 = -2.732446x10-6 C/m2

Hence net charge density on outer surface = +6.37x10-6 -2.732446x10-6 = 3.63755x10-6 C/m2

Part b

Electric field strength = kQ/r2 where Q is total charge inside and r is outer radius.

Charge on sphere = (6.37x10-6)(*(0.260)2) = + 1.352x10-6 C

Charge placed inside =-0.580x10-6

Hence Q = + 1.352x10-6 -0.580x10-6 = 0.77212x10-6

Electric field strength = kQ/r2 = (9x109)(0.77212x10-6)/(0.260)2 = 102.797x103 N/C

Part c

Electric flux = Q’/0

Q’ = -0.580x10-6

Electric flux = Q/0 = (0.580x10-6)/(8.85x10-12) = 0.065536x106 N m2/C

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