The figure below is a section of a conducting rod of radius R1 = 1.30 mm and len

Question

The figure below is a section of a conducting rod of radius R1 = 1.30 mm and length L = 11.00 m inside a thick-walled coaxial conducting cylindrical shell of radius R2 = 10.0R1 and the (same) length L. The net charge on the rod is Q1 = +3.50 10-12C that on the shell is Q2 = ?2.00Q1.

(a) What is the magnitudeEof the electric field at a radial distance ofr=1.50R2?
N/C

(b) What is the direction of the electric field at that radial distance?
---Select---inwardoutward

(c) What is the magnitudeEof the electric field at a radial distance ofr=2.50R1?
N/C

(d) What is the direction of the electric field at that radial distance?
---Select---inwardoutward

(e) What is the charge on the interior surface of the shell?
C

(f) What is the charge on the exterior surface of the shell?
C

Explanation / Answer

A very long conduction cylindrical rod of length L = 11.00 m and radius R1 =1.30 mm, with cylindrical shell of same length outside it (coaxial with it) having a radius R2 = 10.0 R1.Total charge on cylindrical rod is Q1 = 3.50 x 10-12 C and total charge on shell is Q2 = - 2Q1.

(a) If one draws a cylindrical Gaussian surface outside the conducting shell, the total charge withinthat Gaussian surface is - Q1. Assuming the radius of the surface is r, and it’s length is L, and thesurface is coaxial with the rod and shell, the magnitude of the field is constant at and given radius rand is directed perpendicular to the Gaussian surface.The net flux is EA, where A = 2p rL. By Gauss’ law, qenc = - Q1 = e0F= e0E(2p rL)-> E = - Q1/(2pe0rL). At r = 1.5 R2 = 0.0195 m,

calculate E .

b)If Q1 itself is positive, the field is pointing inward toward the central axis.

(c) At r = 2.50 R1, the charge enclosed is + Q1 and E = Q1/[(2pe0L(2.R1)]

(d) Since the charge enclosed is positive the field is directed outward.

(e) Since there is no field (except locally near the atoms) within the conducting shell, any Gaussiansurface within would have no flux. Therefore, such a surface would have a total charge of zeroPage 1 of 3within it. That is, the charge on the interior surface of the shell plus the + Q1 on the rod would add upto zero. The charge on the interior surface would be - Q1.

(f) Since the total charge on the shell is - 2Q1, the charge on the outer surface must also be - Q1.

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