An infinitely long wire of linear charge density +4nC/m is surrounded by a metal

Question

An infinitely long wire of linear charge density +4nC/m is surrounded by a metallic shell of inner radius 4 cm and outer radius 5 cm. The metallic shell has a net linear charge density of +5nC/m .

Part A

What is the linear charge density of the inner surface of the metallic shell?

Give your answer in nC/m^2.

Part B

What is the linear charge density of the outer surface of the metallic shell?

Give your answer in nC/m^2.

Part C

What is the surface charge density of the inner surface of the metallic shell?

Give your answer in nC/m^2.

Part D

What is the surface charge density of the outer surface of the metallic shell?

Give your answer in nC/m^2.

Part E

What is the electric field at 2.0 cm?

Give your answer in N/C.

Part F

What is the electric field at 4.5 cm?

Give your answer in N/C.

Part G

What is the electric field at 6.0 cm?

Give your answer in N/C.

An infinitely long wire of linear charge density +4nC/m is surrounded by a metallic shell of inner radius 4 cm and outer radius 5 cm. The metallic shell has a net linear charge density of +5nC/m . Part A What is the linear charge density of the inner surface of the metallic shell? Give your answer in nC/m^2. Part B What is the linear charge density of the outer surface of the metallic shell? Give your answer in nC/m^2. Part C What is the surface charge density of the inner surface of the metallic shell? Give your answer in nC/m^2. Part D What is the surface charge density of the outer surface of the metallic shell? Give your answer in nC/m^2. Part E What is the electric field at 2.0 cm? Give your answer in N/C. Part F What is the electric field at 4.5 cm? Give your answer in N/C. Part G What is the electric field at 6.0 cm? Give your answer in N/C.

Explanation / Answer

a)-4nC/m^2

b)9nC/m^2

c)Let us take an element of length dx parallel to the axis of the system.

The inner surface will have a charge which will cancel out the charge on the wire.

-4nC*dx=2pi*0.04*sigma*dx

sigma=-15.92nC/m^2

d)Charge on outer surface is 5nC/m more.

Therefore,

4nC*dx+5nC*dx=2pi*0.05*sigma*dx

sigma=28.65nC/m^2

e)E=2k*lamda/d=2*9*10^9*4*10^-9/0.02=3600N/C

f)E=0 (since field inside a conductor is zero. This can also be understood by the fact that the net charege enclosed by a gaussian surface at that distance is 0. So flux and field are 0)

g)E=2k*lamda/d=2*9*10^9*9*10^-9/0.06=2700N/C

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