Question

In: Physics

The figure below is a section of a conducting rod of radius R1=1.30mm and length L=11.00m...

The figure below is a section of a conducting rod of radius R1=1.30mm and length L=11.00m 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=-4.3010−12C that on the shell is Q2=-4.00Q1.

a) What is the magnitude E of the electric field at a radial distance of r = 2.50R2?

b) What is the direction of the electric field at the radial distance (inward, outward, or zero)? Give reasons!

c) What is the magnitude E of the electric field at a radial distance of r = 3.60R1?

d) What is the direction of the electric field at that radial distance (inward, outward, or zero)? Give reasons

Solutions

Expert Solution

(a) Consider a Gaussian surface in the form of a cylinder of length L and radius r = 2.50R2, that is, enclosing both the cylindrical shell and the rod. Now charge enclosed by the Gaussian cylinder is given as, Qenc = Q1 + Q2

= (-4.30*10^-12 - 4*4.30*10^-12) C

= -21.5*10^-12 C

Area of the curved surface of the Gaussian cylinder (as the upper and lower circular bases of the cylinder does not contribute towards the flux) is = 2rL

By Gauss's law, E*2rL = Qenc/o

Or, E*2*3.14*2.5*10*1.3*10^-3*11 = 21.5*10^-12/8.85*10^-12

Or, E*2245.1*10^-3 = 2.429

Thus, electric field, E = 1.08 N/C

(b) As the charge is of negative polarity, so, the electric field will act radially inward.

(c) Cosider as in (a) a Gaussian cylinder of length L and radius r = 3.60R1, that is, this time the Gaussian cylinder encloses only the inner rod.

So, Qenc = Q1 = -4.30*10^-12 C

As before, by Gauss's law, E*2rL = Qenc/o

Or, E*2*3.14*3.6*1.3*10^-3*11

= 4.30*10^-12/8.85*10^-12

Or, E*323.2944*10^-3 = 0.48587

Thus, electric field, E = 0.48587/323.2944*10^-3

   = 1.51 N/C

(d) As the charge on the rod is negative, so, electric field will direct radially inward.


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