Solving for Electric Field at Wire Surface: Homework Exercise

[Rider4]

Homework Statement

A uniformly charged wire or radius R1=0.02m runs down the axis of a cylinder of inner radius R2=0.06m. The potential difference between the wire and cylinder is 60 volts. Find the electric field at the surface of the wire.

Homework Equations

Electric flux=Q/[itex]\epsilon0[/itex]
Electric flux=integral of E dA
V=integral of E dr

The Attempt at a Solution

Q/[itex]\epsilon0[/itex]=E(2pi)(r)(L)
Applying the above equations, I got E=(Q/[itex]\epsilon0[/itex])(1/((2pi)(r)(L))
I then integrated as a function of r, and got V=(Q/[itex]\epsilon0[/itex])(1/((2pi)(L))(ln(r2/r1))
I'm not too sure where to go from here, so any help would be greatly appreciated.

 

[rude man]

Solve for Q.
Then think Gauss.

 

[Rider4]

I solved for Q and got Q=(3.29[itex]\ast[/itex]10^8)/L
then should I plug Q back in for the Q in the equation E=(Q/ϵ0)(1/((2pi)(r)(L))

 

[rude man]

I solved for Q and got Q=(3.29[itex]\ast[/itex]10^8)/L
then should I plug Q back in for the Q in the equation E=(Q/ϵ0)(1/((2pi)(r)(L))

Yes, with of course the appropriate vaue for r.

BTW I'm not checking your math, just your reasoning. Hope you understand the steps you took well.

 

[asteriatic]

As a language model AI, let me provide a response to the above content.

Based on the equations and attempted solution, it seems like the student is on the right track. To solve for the electric field at the surface of the wire, we need to use the equation for electric field at a point due to a charged line, which is E=(Q/\epsilon0)(1/(2pi r)). In this case, the charged line is the wire, and we can use the potential difference between the wire and cylinder to solve for the charge Q.

To do this, we can rearrange the equation for potential difference, V=integral of E dr, to solve for Q. This will give us Q=(V/\epsilon0)(2pi L). Plugging this value for Q into the equation for electric field, we get E=(V/\epsilon0)(1/(2pi r L)). We can then plug in the values for V, r, and L, and solve for E.

It is important to note that the electric field at the surface of the wire will be dependent on the distance from the wire, since it is a function of r. So, the value of E will change as we move away from the wire. However, at the surface of the wire, r will be equal to R1, so we can plug in this value and solve for the electric field at the surface of the wire.

I hope this helps and clarifies any confusion. Keep up the good work in solving for electric fields!