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Calculating electric field strength, potential difference, and force.

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data
    esou.gif

    I'm pretty sure I have done questions 1 to 4 correctly, it is just question 5 I am struggling with.
    I have an exam in one week and this is revision so it will not get marked by a tutor, and it would really help if someone could give my answers a quick check and help me with question 5.
    Thanks!

    1)
    a) Calculate the electric field strength at position (2,3,0) of a 1 coulomb point charge placed at position (0,0,0)

    2)
    a) Calculate the electrical potential at point (2,3,0)
    b) Calculate the electrical potential at point (0,0,0)
    c) Calculate the electrical potential difference
    e) Calculate the electrical potential difference between (x,y,z) and (x1,y1,z1)

    3)
    a) A 0.3 Coulomb point charge is placed at (2,3,0). Calculate the force on this charge

    4)
    a) Calculate the electrical potential energy of this point charge.

    5)
    a) If the charges mentioned were not point charges, how would the equations and answers change.
    b) Derive the equations needed for these questions if the charges where not point charges, but instead:

    i) Infinitely long cylindrical parallel wires.
    ii) Infinitely long X mm thick parallel micro strip wires
    ii) An infinitely long co-axle cable with the inner conductor at charge q and the outer at charge Q
    iii) Two infinitely large parallel plates

    c) How would these equations change if the objects where of a finite length.

    2. Relevant equations

    1)
    [itex]\underline{F} = \frac{qQ}{\underline{r}^2} * \frac{1}{4\pi\epsilon_{0}}[/itex] * [itex]\underline{\widehat{r}}_{qQ}[/itex]

    [itex]\underline{\widehat{r}}_{qQ}[/itex] Is the unit vector in the direction to get from q to Q

    2)
    [itex]\underline{E} = \frac{Q}{\underline{r}^2} * \frac{1}{4\pi\epsilon_{0}}[/itex] * [itex]\underline{\widehat{r}}[/itex]

    3)
    [itex]\underline{F} = \underline{E}q[/itex]

    4)
    [itex]\underline{V} = - \oint\underline{E}.dl [/itex]


    3. The attempt at a solution

    1a)
    r=SQRT(2^2+3^2) = 3.6
    direction of r is pointing perpendicularly away from the point charge.
    r unit vector = (2i+3j)/3.6
    E=1/(3.6^2) * 8.99*10^9 * (2i+3j)/3.6
    E = 6.94*10^8 * (2i+3j)/3.6
    E =(3.85*10^8)i + (5.78*10^8)j

    2a)
    V = E * 3.6
    V = (1.39*10^9)i + (2.08*10^9)j

    2b)
    V = 0

    2c)
    V = (1.39*10^9)i + (2.08*10^9)j

    2e)
    will be done exactly the same but with letters instead of numbers.

    3a)
    F = (3.85*10^8)i + (5.78*10^8)j * 0.3
    F= (1.155*10^8)i + (1.734*10^8)j

    4a)

    Energy = (1.155*10^8)i + (1.734*10^8)j * 3.6
    Energy = (4.158*10^8)i + (6.2424*10^8)j
    Energy = SQRT(a^2+b^2)
    Energy = 7.50*10^8 J

    (should I have integrated the force across the distance instead of just multiplying them?)

    5a) I have absolutely no clue here. I think I need to derive the equations I gave in the beginning from a more general form. Probably from maxwell's equations. can someone please help me out on this bit?

    Thank you!
     
  2. jcsd
  3. Mar 19, 2013 #2

    BruceW

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    Homework Helper

    1)a) good, you've got this correct.
    2)a) This is not right. you have used [itex]\vec{E}r[/itex] But this does not equal the electric potential. You are close, though. What should you use instead of [itex]\vec{E}[/itex] ?
    2)b) I don't think your answer for this is correct. Close to (0,0,0) the electric potential becomes infinitely great, and at (0,0,0) it is not even defined. It is not zero.
    2)c) Given that I think the answer to 2)b) should be infinite/not defined, then this question does not make sense... Are you sure you wrote down the question correctly?
    2)e) What happened to 2)d) haha? Um, for the question, again you've not used the correct definition for the electric potential.
    3)a) Yes, this is correct. You have correctly used your answer from part 1)a)
    4)a) again, it is not right because the electric potential is not [itex]\vec{E} r [/itex]
     
  4. Mar 19, 2013 #3
    Ah yes I have wrote down the co-ordinates for question 2b wrong, it should be (1,0,0). I'l do this again in a minuite and then post my answer. And for qusetion 2a) do I use D instead of E, where D=E*permitivity of free space?
    It is suprisingly difficult to find the equation for electrical potential online.
     
  5. Mar 19, 2013 #4
    2a)
    v= (1.39*10^9)i + (2.08*10^9)j * 8.85*10^-12
    v= (0.018408)j+(0.0123015 i)

    2b)
    first find E at (1,0,0)
    r=SQRT(1^2+0^2) = 1
    direction of r is pointing perpendicularly away from the point charge.
    r unit vector = (1i+0j)/1 = i
    E=1/(1^2) * 8.99*10^9 * i
    E = (8.99*10^9)i + (0)j
    V= E * permitivity of free space * 1
    V = (0.0795991485018)i + (0)j

    2c)
    electrical potential diffrence equals:

    V = v1 - v2

    V = (0.018408)j - (0.0672976)i

    question 2d) was asking for the potential at (x,y,z)
    this and question 2e) should now be similar to b and c now that I am using the correct deffinition for electrical potential.

    if question 3a) is correct shouldn't question 4a) also be correct? I have just used E=F*r

    Thanks for replying by the way, it is much appreciated.
     
  6. Mar 19, 2013 #5

    BruceW

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    Homework Helper

    No, that's not it. And yeah, it is harder than I expected to find a good, simple explanation of the electric potential by searching online. Anyway, I am looking at your 'relevant equations' section, and now I see maybe where the problem is. You have written:
    [tex]\underline{V} = - \oint\underline{E}.dl[/tex]
    But on the left-hand-side, the V should not have an underline. The electric potential is not a vector at all. It is a scalar quantity (like temperature is a scalar too). Also, on the left-hand-side, it should be the difference in electric potential. The right-hand-side is correct, it is the line integral of the electric field. (EDIT: actually, it is correct, but there should be an underline on the 'dl' since it is a vector). So the electric field is a vector, but it does an inner product with the line element, so on both sides of the equation are scalars.

    And as it happens, the electric potential due to a single point charge is very simple, and has a nice relation to the electric field due to a point charge. You were using [itex]\underline{V}=\underline{E}r[/itex] Which is close, but not quite right because the electric potential is a scalar. So keeping this in mind, the right-hand-side should be a scalar too. So have a guess, what is the actual equation? p.s. the electric potential due to a single point charge is called the Coulomb potential. You've probably seen it before, but maybe forgot the exact equation.

    Edit: I think the (vector) dl is more often called the differential vector, not the 'line element' as I have called it. This page is good: http://en.wikipedia.org/wiki/Line_integral
    Also, you don't need to worry too much about line integrals of a vector. Because for a single point charge, the electric potential happens to be fairly simple. You just need to use the equation for it (if you can remember it)
     
    Last edited: Mar 19, 2013
  7. Mar 19, 2013 #6
    Ahhh okay electrical potential is a scalar, this has cleared a few things up. I assumed it was a vector because voltage can be negative (so it sort of has a direction). I now see that this is just becuase voltage is a difference in electrical potential, and that even a difference in temperature could be negative.

    So the actual equation will be:
    volatge = the dot product (scalar product) of E and r. This means (the amount of E in the direction of r) multiplied by r.
    V= Ecos(theta)r
    theta = 0
    V=|E|*r

    if |E| is (Q/r^2) * 8.99*10^9

    I end up with

    V = 8.99*10^9 * Q/r
    V = 8.99*10^9 * 1/3.6
    V = 2.50 *10^9
     
    Last edited: Mar 19, 2013
  8. Mar 19, 2013 #7
    By the way 8.99*10^9 is coming from 1/(4pi*epsilonNought)
     
  9. Mar 19, 2013 #8

    BruceW

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    V=|E|*r is the correct equation. But it doesn't really come from using the dot product between the vectors E and r. It is just a 'lucky coincidence' that for a single point charge V=|E|*r

    You can derive the equation by integrating the electric field along a differential vector from infinity to r. This is really where the equation V=|E|*r comes from. This is maybe more like undergraduate-level physics though.

    Yep! That is correct, for question 2)a)
     
  10. Mar 19, 2013 #9

    BruceW

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    I'm glad I could have helped!

    Yeah, I guessed that :)
     
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