A cylindrical insulator with a radius of R = 0.150 m and length of 3.00 m carrie

Question

A cylindrical insulator with a radius of R = 0.150 m and length of 3.00 m carries a uniform charge density equal lo 4.00 mu C/m1. We want to find the electric field at a point r = 0.25 from the center of the cylinder What is the correct area of the Gaussian surface for this case? (circle ONE equation only) 4pir^2 4pR^2 4/3piR^3 pir^2L piR^2 L 2pirL 2piR L What is the correct charge inside the Gaussian surface for this case?(circle ONE equation only) 4PIR^2 4PI R^2 4/3PIR^3 PIR^2L PIR^2 L 2PIRL 2PIR L Use the above information to calculate the electric field on the Gaussian surface?

Explanation / Answer

Radius of the cylinder R = 0.15 m

Length of the cylinder L = 3 m

Charge density = 4 µ C/ m3 = 4 * 10-6 C / m3

Now we need to find the electric field at a point P at a distance r = 0.25 m.

So we consider a cylindrical Gaussian surface of radius r and length L

From gauss law we can write ,

E.ds = Qen / 0

Qen = * * R2 * L

Electric flux through the circular faces is zero. So from gauss law ,

E.ds = * * R2 * L / 0

E ds = * * R2 * L / 0

Then E * ( 2 * r * L ) = * * R2 * L / 0

So     E = [ * R2 / ( 2 * 0 * r ) ]

This is the case when r > R

E = ( 4 * 10-6 * 0.15 * 0.15 ) / ( 2 * 8.85 * 10-12 * 0.25 )

    = 0.02034 * 106 V/m

    = 20340 V/m

So the correct options are

a) the area of the Gaussian surface is S = 2 r L

b)the charge enclosed by the gaussian surface is Qen = * * R2 * L

c)the electric field is E = 20340 V/m

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