Electric Field through an infinitely long hollow cylinder?

Question

Infinitely long hollow cylinder of radius .03 m is charged with a uniform surface charge of .8 uC/m^2. Use Gauss's law to find the magnitude and direction of the field at a distance of .015 m and .06 m from the center of the cylinder

Answer 1

Hello Tim, as one point is within the charged cylinder the field is 0. Since the flux is due to the charge enclosed by the Gaussian surface. But there is no charge inside the imaginary surface with radius ie half of the given ie 0.015 m

Okay for the second case. Let us imagine a cylinder of radius 0.06 m with some length L

So surface area of the charged cylinder covered is 2 pi (0.03) L

So the charge for this area = 2 pi (0.03) L (0.8) * 10^-6 C

Hence q /Epsilon Not is the flux coming out of our imaginary Gaussian surface.

Let E be the field (uniform) to be found out.

So be calculation the flux Integral E ds

Or E * 2 pi (0.06) L

By Gauss law , q / epsilon not = E * 2 pi (0.06) L

or E * epsilon not = [2 pi (0.03) L (0.8) * 10^-6 ] / [2 pi ( 0.06) L] = 0.4*10^-6

Required E = 0.4*10^-6 / 8.85*10^-12 = 4.52 * 10^4 V m^-1

Answer 2

Electric Field Inside A Cylinder

Answer 3

we consider the hollow cylinder as a straight wire and we construct the gaussian surface .015 m away from the cylinder ,the result of electric field due to charged straight wire can be used ie....

E=λ/2πrε0

E=0.8 X 10^ -6 / 2 X 3.14 X 0.015 X 8.854 X10^ -12

E=9.6 X 10^5 N/C

the direction is radial

similarly for .06m

E=λ/2πrε0

E=0.8 X 10^ -6 / 2 X 3.14 X 0.06 X 8.854 X10^ -12

E=2.4024 X 10^5 N/C

the direction is radial

Answer 4

a million) C. the electric powered powered powered potential is inversely proportional to the radius of the sector. 2) F = ok q1q2 / a million^2 = ok q1q2 ... Fnew = ok q1q2 / 2^2 = kq1q2 / 4 = F/4 ----- (C) 3) mgh = 10 x 9.8 x 10 = 980 J