Infinite Line of Charge and Electric Potential Difference

[tomlichu]

Homework Statement

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?

Homework 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=V_B-V_A

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=V_B-V_A=-∫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$\cdot∑$q_n/r_n)
Would the sum of all the point charges just be λ since q/l is just charge over lenght?

 

[Simon Bridge]

Welcome to PF;

I got
E=λ/(2∏ε0r) when I took the integral of ∫2∏rE⋅dl from 0 to length L and set it equal to Q/ε0

... why did you pick length L?
I don't follow your reasoning.

 

[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
I used the website above to help me in my calculations.

 

[Simon Bridge]

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.

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

According to your problem statement,

1. Consider a line of charge (with λ charge per unit length which extends along the x axis from x=-∞ to x=0

... 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

... any point along the positive x-axis

... 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.

 

[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.

 

[tomlichu]

Ahhh I see this makes more sense

Can you see why there is a cube in the denominator?

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

 

[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=(X_f-X_i) if I was to integrate in terms of dx?

 

[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.

Wouldn't r=(Xf-Xi) if I was to integrate in terms of dx?

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.