A hollow, conducting sphere with an outer radius of 0.252 m and an inner radius

Question

A hollow, conducting sphere with an outer radius of 0.252 m and an inner radius of 0.196 m has a uniform surface charge density of +6.85 times 10^-6 C/m^2. A charge of-0.720 mu C is now introduced into the cavity inside the sphere. What is the new charge density on the outside of the sphere? sigma = -2.1723 middot 10^-6 C/m^2 Calculate the strength of the electric field just outside the sphere. E = N/C What is the electric flux through a spherical surface inside the inner surface of the sphere? phi = N middot m^2/C

Explanation / Answer

Given,

ro = 0.252 m ; ri = 0.196 m ; sigma = 6.85 x 10^-8 C/m^2 ; q = -0.720 uC

A)The charge now on the outer surface need to be known

Q = 6.85 x 10^-6 x 4 x pi x 0.252^2 = 5.46 x 10^-6

Qn = 5.46 x 10^-6 - 0.72 x 10^-6 = 4.74 x 10^-6 C

sigma(new) = 4.74 x 10^-6/4 pi x 0.252^2 = 5.94 x 10^-6

Hence, sigma = 5.94 x 10^-6 C/m^2

b)E = Q/4 pi e0 ro^2

E = 5.94 x 10^-6/(4 x 3.14 x 8.85 x 10^-12 x 0.252^2) = 8.41 x 10^5 N/C

Hence, E = 8.41 x 10^5 N/C

c)Phi = Q/e0

phi = 0.72 x 10^-6/(8.85 x 10^-12) = 8.14 x 10^4 Nm^2/C

Hence, 8.14 x 10^4 Nm^2/C

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