A thin, metallic, spherical shell of radius a = 6.0 cm has a charge q a = 6.00×1

Question

A thin, metallic, spherical shell of radius a = 6.0 cm has a charge qa = 6.00×10-6 C. Concentric with it is another thin, metallic, spherical shell of radius b = 10.20 cm and charge qb = 4.00×10-6 C. Find the electric field at radial points r where r = 0.0 cm.

Find the electric field at radial points r where r = 8.1 cm.

Find the electric field at radial points r where r = 15.3 cm.

Discuss the criterion one would use to determine how the charges are distributed on the inner and outer surface of the shells. What is the charge on the outer surface of the outer shell?

Explanation / Answer

At r = 0.0

According to gauss's law , Electric field inside a sphere is 0 . So electric field at r = 0.0 is 0

At r = 8.1 cm (between the two spheres)

Electric field is only due to the smaller spherical shell , which is given as

E = k qa / r2 = (9 x 109) (6 x 10-6) / (8.1 x 10-2)2 = 8.23 x 106 N/C

At r = 15.3 cm

Electric field is due to the smaller spherical shell and larger shell, which is given as

E = k qa/r2 + k qb/r2 = k (qa + qb) / r2 = (9 x 109) (6 x 10-6 + 4 x 10-6) / (0.153)2 = 3.85 x 106 N/C

On the outer surface of outer sphere net charge is given as

Q = qa - qb = (6 x 10-6 - 4 x 10-6) = 2 x 10-6 C

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