# Infinite Line of Charge and Electric Potential Difference

1. Apr 19, 2014

### tomlichu

1. The problem statement, all variables and given/known data
1. Consider a line of charge (with λ charge per unit length which extends along the x axis from x=-∞ to x=0
(a) Find all components of the electric field vector at any point along the positive x-axis

(b) Find the electric potential difference between any point on the positive x-axis and x=1 m

(c) If λ=0.1 μC per meter and a proton is placed at x=1 m with zero initial speed, does the proton's speed ever reach 106 m/s. If so, where does it reach this speed?

2. Relevant equations

Charge of a proton=1.602$\cdot$10-19 C
Mass of a proton=1.673$\cdot$10-27 kg
Gauss' Law Flux=∫E$\cdot$dA=Q/ε0
Q=λ$\cdot$L
ΔV=VB-VA

3. The attempt at a solution
For part a, I got
E=λ/(2∏ε0r) when I took the integral of ∫2∏rE$\cdot$dl from 0 to length L and set it equal to Q/ε0

For part b, not sure if this is correct, but I used the equation
ΔV=VB-VA=-∫E$\cdot$dR from A to B
I substituted the answer I found in part A for E and got -λ$\cdot$lnX/2∏ε0
I am not sure if this is the correct answer, it sort of makes sense, the further the particle is on the x-axis, the greater the potential

(c) For part c, I presume that since the line of charge is infinitely long, the proton will eventually reach that speed. I tried to find the charge Q using the equation Q=λL, but since this is an infinitely long line, Q increases as L increases. The equation for electrical potential energy is U=q0/(4∏ε0$\cdot∑$qn/rn)
Would the sum of all the point charges just be λ since q/l is just charge over lenght?

Last edited: Apr 19, 2014
2. Apr 19, 2014

### Simon Bridge

Welcome to PF;
... why did you pick length L?

Last edited: Apr 19, 2014
3. Apr 19, 2014

### tomlichu

Since this is an infinitely long line of charge, I calculated the electric field by assuming there is a cylinder wrapped around the line of charge because the electric field is uniform on the surface of the cylinder. I chose length L as an arbitrary length for the length of the cylinder. The line of charge extends on the x axis from x=-∞ to x=0, if that makes sense. I am not sure if that is the correct way
http://faculty.wwu.edu/vawter/PhysicsNet/Topics/ElectricForce/LineChargeDer.html [Broken]
I used the website above to help me in my calculations.

Last edited by a moderator: May 6, 2017
4. Apr 20, 2014

### Simon Bridge

It is important to sketch out the situations you have to do maths for.

... that puts the line of charge along the negative x axis. Which you noticed.

You are then asked to find the electric field vector at
... sketch this out.
Any point on the surface of a cylinder about the x axis cannot be on the positive x axis.
The example you are copying was calculating the charge on the y axis - a totally different situation.

Last edited: Apr 20, 2014
5. Apr 20, 2014

### Simon Bridge

What I like to do is construct the integral

The electric field at position $\vec r$ due to a charge $dq$ at position $\vec r'$ is given by: $$d\vec E = \frac{dq}{4\pi\epsilon_0|\vec r-\vec r'|^3}(\vec r-\vec r')$$ ... you should recognize this as the usual equation for the electric field due to a point charge that is not at the origin.
Can you see why there is a cube in the denominator?

For the total field, you just add up the field due to all the charges. In the case of a continuous charge distribution, this means integrating over the region containing the charge.

In your case, the vectors have simple relationships.

So start by drawing a sketch.

Using your sketch, and your knowledge of the definition of the electric field vector...
Anywhere on the positive x axis, which direction does the electric field point in?

Write down the expression for the electric field $d\vec E$ at $\vec r = (x,0,0)=x\hat\imath :x>0$ due to the charge element $dq$ at position $\vec r' = (x',0,0) = x'\hat\imath:x'<0$.

Then express $dq$ in terms of the small length along the negative x-axis $dx'$.

Now you should have an integral to do.

Last edited: Apr 20, 2014
6. Apr 20, 2014

### tomlichu

Ahhh I see this makes more sense

So the equation you are using is dE=dq/(4∏r2)

You made a cube because you pulled out a (r-r'). Did you pull it out so so it becomes a vector quantity?

And anywhere on the x-axis, the vector field point parallel to the x axis in the positive direction I believe

I really appreciate the help you have given me so far, electricity in Physics is not my strength

7. Apr 21, 2014

### tomlichu

Drawing a diagram really helps
So I would set dq=λdx
and then dE=dq/4∏ε0r2, and I would substitue dq for λdx and integrate both sides because the change in charge q over the change in distance x is equal to the charge density.
Is this correct direction I'm heading towards? I still do not completely understand why you made the denominator a cube though. Wouldn't r=(Xf-Xi) if I was to integrate in terms of dx?

8. Apr 21, 2014

### Simon Bridge

You are thinking in the right direction.

Notice though - your version of the equation dE=dq/4∏ε0r2 is a vector on the LHS but not on the RHS - and you have an r where you should have some x's.
That's the idea.
Xf is the place you want to calculate the field and Xi is the location of the charge element - so you'll have to integrate over Xi and dq=λdXi you follow?

Compare with mine ;) making the RHS a vector is the reason there is a cube in the denominator of my version: $\hat r = \vec r/r$.

So you can say:
dE=(dq/4∏ε0r2)(r/r) ... see the extra r in the denominator?

or dE=dq/4∏ε0r2

see?

The second one is where you are headed - to use it you need to know which direction dE will point in by another means. You can get there by looking at the symmetry.