A thin-walled metal spherical shell of radius a = 3.0 cm has a charge qa = 7.00×

Question

A thin-walled metal spherical shell of radius a = 3.0 cm has a charge qa = 7.00×10-6 C. Concentric with it is a thin-walled metal spherical shell of radius b = 6.90 cm and charge qb = 9.00×10-6 C. Find the electric field at distance r = 0.0 cm from the common center. Find the electric field at distance r = 4.9 cm from the common center. Find the electric field at distance r = 10.3 cm from the common center. 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

PART 1

Electric field at r =4.9cm (a<r<b) = 4.9*10-2

This point is in between the spherical shells.outside the inner shell and inside the outer shell.

at this point electric field is due to only inner shell as electric field due to outer shell is zero.

E = K*qa/r^2

E = 9.10^9*7*10-6/ (4.9*10-2)2

E = 26.24*106 N/C

PART 2

r= 10.3cm

This point is outside of both the sphere so at this point field is due to both shells.

E = k*qa/r^2 +K*qb/r^2

E = K/r^2[qa+qb]

E = 9*10^9/(10.3*10.3*10-4)[7*10^-6 +9*10^-6]

E = 13.57*106 N/C

PART 3

We can use conservation of charge to find chrge on inner and outer surface.

Q = qa+qb

Q = 16*10-6 C

Charge on the outer surface of outer shell is the sum of charges on inner shell and

outer shell.

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