A hollow, conducting sphere with an outer radius of R 1 = 0.270 m and an inner r

Question

A hollow, conducting sphere with an outer radius of R1 = 0.270 m and an inner radius of  R2 = 0.210 m has a uniform surface charge density of = +6.67 × 106 C/m2. A charge of  Q = -0.600 C is now introduced into the cavity inside the sphere.

1. What is the new charge density on the outer surface of the conducting sphere?

2. Use Gauss's law to calculate the strength of the electric feild at a distance 0.5 m from the center of the sphere.

3. What is the electric flux through a spherical surface with radius 9*10^-2 m with the hollow conductor?

4. What is the strength of the electric feild at a point 9*10^-2 m from the the center of the spherical conductor?

5. In what direction is the electric feild found in part D and why?

Please explain steps!!!!

Explanation / Answer

1) charge on outer surface =6.67e-6*4pi*0.27^2 - 0.6e-6

= 5.5103e-6

Charge density on outer surface = 5.5103e-6*(4pi*0.27^2)

= 5.05*10^-6 C/m^2

B) By Gauss law, E = kq_enclosed /r^2

= 9e9*5.51e-6/0.5^2 = 198360 N/C

C) Flux = Qenclosed/e0 = 0.6e-6 /8.85e-12

= 67797 Nm^2/C

D) E = flux/area = 67797/(4pi*0.09^2) = 666063 N/C

E) towards the center because charge enclosed is negative

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