Question

A long, conductive cylinder of radius R1 = 2.45 cm and uniform charge per unit length λ = 302 pC/m is coaxial with a long, cylindrical, non-conducting shell of inner and outer radii R2 = 8.57 cm and R3 = 9.80 cm, respectively. If the cylindrical shell carries a uniform charge density of ρ = 79.8 pC/m^3, find the magnitude of the electric field at the following radial distances from the central axis:

1.74 c.m= _________ N/C

6.25 c.m= __________ N/C

9.07 c.m= ___________N/C

10.5 c.m= ____________ N/C

Answer 1

Hi,

Hope you are doing well.

This question is a simple application of Gauss's Law.

GAUSS'S LAW:

Statement: The flux of the net electric field through a closed surface equals the net charge enclosed by the surface divided by  

First of all we have to find the expression for finding electric field produced by conducting cylinder as well as the charged shell(non-conducting), placed coaxial to it using Gauss's law.

(1)Conducting cylinder

Considering a gaussian surface in the form of a cylinder  r > R1, the electric field has the same magnitude at every point of the cylinder and is directed outward. The electric flux is then just the electric field times the area of the cylinder. (Electric flux at lids of cylinder become zero since area and electric field is perpendicular)

For  

(2)Charged cylindrical shell.

Considering a gaussian surface in the form of a cylinder  r > R3, the electric field has the same magnitude at every point of the shell and is directed outward. The electric flux is then just the electric field times the area of the shell. (Electric flux at lids of shell become zero since area and electric field is perpendicular)

For  

Now lets move through each cases:

(1) 1.74 cm= _________ N/C

Given that R1= 2.45 cm

ie; 1.74 cm<R1

Electric field at 1.74 cm, E=0

(2) 6.25 cm= __________ N/C

Given that inner radius of the shell R2=8.57 cm.

ie; 6.25 cm<R2

There is no electric field due to shell at 6.25 cm. (Same reason as of conducting cylinder)

Electric field due to Conducting cylinder at 6.25 cm,

(3) 9.07 cm= ___________N/C

Given that inner radius of the shell R2=8.57 cm.

ie; 9.07 cm>R2

There is no electric field due to shell at 9.07 cm. (Same reason as of conducting cylinder)

Electric field due to Conducting cylinder at 9.07 cm,

(4) 10.5 cm= ___________N/C

Given that outer radius of the shell R3=9.80 cm.

ie; 10.5cm>R3

There will be electric field due to shell as well as conducting cylinder at 10.5 cm.

So we have to find the total flux at 10.5 cm.

Hope this helped for your studies. Keep learning. Have a good day.

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