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## Homework Statement

Find the magnitude of the electric field intensity at a distance 0.15m from the axis of a long, straight conductor, which has radius 0.03m, and a charge density of 4.0 C/m.

## Homework Equations

Eflux = E.A

Eflux = q/e0

q = sigma*h

## The Attempt at a Solution

Eflux = E*Area

Thus, Eflux = E*2pi*r*h (Surface area of the gaussian cylinder we superimpose over the conductor)

Also, Eflux = q/e0 -> where e0 is the permittivity of free space and q is the enclosed charge in the gaussian surface.

q = sigma*h

Therefore we substitute for the q, and equate the two Eflux equations to get:

Eflux = E*2pi*r*h = sigma*h/e0

Solve for E and we get:

E = sigma/(2pi*r*e0)

Now we substitute the given values:

4/(2pi*(0.03)*(8.85*10^-12)) = 2.40*E12 or 2.40*10^12 N/C

I have no idea how to find the electric field at the distance 0.15m away from the surface. Please help! I would need the value of q to do this, however this is not a point charge we're dealing with but a surface charge, so I'm a bit lost.