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Infinite Line of Charge and Electric Potential Difference

  1. Apr 19, 2014 #1
    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[itex]\cdot[/itex]10-19 C
    Mass of a proton=1.673[itex]\cdot[/itex]10-27 kg
    Gauss' Law Flux=∫E[itex]\cdot[/itex]dA=Q/ε0
    Q=λ[itex]\cdot[/itex]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[itex]\cdot[/itex]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[itex]\cdot[/itex]dR from A to B
    I substituted the answer I found in part A for E and got -λ[itex]\cdot[/itex]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[itex]\cdot∑[/itex]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. jcsd
  3. Apr 19, 2014 #2

    Simon Bridge

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    Welcome to PF;
    ... why did you pick length L?
    I don't follow your reasoning.
     
    Last edited: Apr 19, 2014
  4. Apr 19, 2014 #3
    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
  5. Apr 20, 2014 #4

    Simon Bridge

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

    According to your problem statement,
    ... 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
  6. Apr 20, 2014 #5

    Simon Bridge

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    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
  7. Apr 20, 2014 #6
    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 :smile:
     
  8. Apr 21, 2014 #7
    Drawing a diagram really helps :smile:
    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?
     
  9. Apr 21, 2014 #8

    Simon Bridge

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    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.
     
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