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Homework Help: Electric potential on circle problem

  1. May 16, 2010 #1
    [PLAIN]http://img526.imageshack.us/img526/5079/29693720.jpg [Broken]
    im struggling to start this question
    do i calculate all of the potentials seperatly and add them up?
    and to calculate the potential on the circle and 1/4 circle would i just use 2pi r ?
    and for the 1/4 cirlce definite intergrate it from pi/2 to pi
    how would i go about the diapole ?
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. May 17, 2010 #2

    AEM

    User Avatar

    Yes, you calculate each potential separately and then add them all up. For the curved pieces construct an integral for a potential dV from a bit of charge dq. then integrate. Your limits of integration depend on how you choose to draw your diagram, but the total angle is the same no matter how you set it up. For the dipole, do each charge separately.
     
  4. May 19, 2010 #3
    would i be able to have a hand with the intergral
    im kind of struggling to work it out
    so the potential from one point on the big circle is k(-3x10^-6 / 0.06)
    where k = 8.99x10^9
    so would i times this by the intergral of 2pir?
    but that just turns into the area of a circle
    coudl i just times that by the perimeter or the circle?
    and for the smaller curve could i times its charge by 1/4 2pi r
     
  5. May 19, 2010 #4
    Let's start with the potential due to a point charge.

    [tex]
    \begin{flalign*}
    V & = & & k \ \frac{q}{r} \ \mbox{ or } \ \frac{1}{4 \pi \epsilon_0} \ \frac{q}{r}\\
    \end{flalign*}
    [/tex]

    We want to break the ring into infinitesimal segments and find the potential due to one of these segments right? Then we can add them all up using the integral. So we want to calculate V=[tex]\int dV[/tex]. What is the equation for dV?
     
  6. May 19, 2010 #5
    k * intergral (dq/r)
    is that right?
    how do i relate it to the perimeter of the circle? cause it doesnt say how thick the wire is
     
  7. May 19, 2010 #6
    Yes that's right. Now how else you can express dq? There is a total charge Q spread uniformly on the ring and you know the size of the ring, since you are given the radius.

    When a problem states an object is "thin" they mean it has no thickness. So imagine the ring to be a line bent into a circle of radius R.
     
  8. May 19, 2010 #7
    ok so is the total charge Q = -3 microC ?
    so it woudl be like

    k* (Q(2pir)dr)/root(z^2 + r^2) ?
    thats what they get in the book, but im just abit confused
    why is it z^2 + r^2? it doesnt make sense with phythagorus unless im thinking silly
     
  9. May 19, 2010 #8
    Let's start by defining our coordinate system. The simplest case would be to place the ring in the xy plane, and center it on the z-axis. Now if we do this, then according to our diagram we are evaluating the potential at the origin of our coordinate system correct?
    So in our new equation for the potential,
    [tex]

    \begin{flalign*}
    dV & = & & \ \frac{1}{4 \pi \epsilon_0} \ \frac{dq}{r}\\
    \end{flalign*}

    [/tex], r is the distance from the differential segment to the point where we are calculating the potential so, in our case this is the radius since we want to evaluate the potential at the origin. Now think about how we can express dq. We want to integrate over all the segments of the ring right? So we need for dq to somehow be expressed in terms of some differential length. We know the total charge on the ring and we know how long it is so we can define a linear charge density as [tex]\lambda = Q/l[/tex] where l is the length of the ring and thus has units of charge/length. Then dq is?
     
  10. May 19, 2010 #9
    ohh i see
    so dq is lamda dx
    then because its a constant you can pull it out of the intergral?
    and it looks like

    k lambda (ingtergral of(dx/r))
    is that right?
    the next step puts r = root(x^2 + d^2)
    is r always = to somthing like that?
    is that like a rule ive missed?
     
  11. May 19, 2010 #10
    That's correct, although I would write it as dl, so you don't get confused with integrating over the x coordinate. So this is just the line integral around the loop which is just equal to the perimeter.

    This expression for r is for the more general case. Suppose we have the ring lying in the xy plane, centered on the z-axis as before, but now we want to evaluate the potential due to the ring at some point on the z-axis a distance z above the origin. Remember r in our equation for the potential is the distance from the element dq to the point at which we're evaluating the potential. Then r is no longer just equal the radius of the ring since we're not finding the potential at the origin, instead it is the distance from the element dq to our point of evaluation somewhere on the z-axis above the ring. So if you draw this out, you can use the pythagorean theorem to find the distance r. Do you see how they get sqrt(z^2 + R^2)? To avoid confusion I am using R as the radius of the ring. r is still the distance from dq to the point where we are evaluating the potential.
     
    Last edited: May 19, 2010
  12. May 19, 2010 #11
    ohhh right right right
    i get it now
    sorry i was reading two different examples
    one was a disc and the other was a line
    the disc one was d
    but i udnerstand now

    so now i juts intergrate with the limits L and 0 right?
    and thatll give me the charge for the cirlce?

    ok i think i got that part down, and with the next part, i juts do somthing similar with the length of the curve and a new distance?
    so i'd add them together?
    and ill just have a look at how to work with the dipole
     
  13. May 19, 2010 #12
    sorry i have to go out for a bit
    thanks for your help so far its been really good
    im just looking at this disc example
    how come we wouldnt use
    dq = sigma 2pir dr
    cause the line example looks like its finding the charge on a line that gets further and further away from the point
    but the ring stays the same distance from the point
    does that matter?
     
  14. May 19, 2010 #13
    aw man
    i intenesly over think these questions
    but ive been thinking
    and this is what made me think what i said before
    like because it looks like by this pictre the intergral is finding the charges at different distances from the point
    and the ring has a constant distance from the point
    i dono i hope my blabber makes some kind of sense
    thanks for bearing with me
    [PLAIN]http://img168.imageshack.us/img168/9555/26440124.jpg [Broken]
    does it add it up with a different r every time? like a different distance from the point?
     
    Last edited by a moderator: May 4, 2017
  15. May 19, 2010 #14
    That is what you would use. Why do you think you wouldn't be able to write dq this way?

    You're right, we are finding the potential due to each infinitesimal charge dq and then adding them all up. What we end up doing is expressing the charge as dq= [tex]\lambda dx[/tex] so that we end up integrating over the coordinate x, or in the case of a surface charge density the area da, etc.. This is allows us to account for the geometry of the problem. In our equation [tex]
    \begin{flalign*}
    dV & = & & \ \frac{1}{4 \pi \epsilon_0} \ \frac{dq}{r}\\
    \end{flalign*}
    [/tex] for the potential, r is the distance from the point dq to the point at which we're evaluating the potential. So in the ring problem, this is just equal to R, the radius of the ring since we were evaluating the potential at the origin. Every dq is the same distance R from the origin so we just take r=R=constant and pull it out of the integral. Now in the case of the disc, r is now equal to sqrt(x[tex]^{2}[/tex] + d[tex]^{2}[/tex]) as you can see from the diagram. And we're integrating over dx, so we cannot just pull r out as a constant in the same way we did for the ring problem. Since r is a function of x, it remains inside the integral.
     
  16. May 19, 2010 #15
    so would we end up getting as the intergral

    (k lamda)/r Intergral(dl)
    r as a constant and out of the intergral?

    so it would end up being L * (k lamda)/r + 0 ? with limits L and 0
     
  17. May 19, 2010 #16
    For the line charge pictured in the diagram that you uploaded r is not constant. Remember r is the distance from dq to the point at which we're evaluating the potential. So for the ring, r is constant and equal to the radius. For the disc, r is not constant. And for the line charge in that diagram r is most certainly not constant. As you move down the line from the origin to L, you're moving further and further away from the point of evaluation, r is changing the whole time. r as shown in the diagram is r=sqrt(x[tex]^{2}[/tex] + d[tex]^{2}[/tex]) which is a function of x. And dq is dq=[tex]\lambda dx[/tex] so we're integrating over x. So like I said before r cannot be pulled out of the integral since it is a function of x. Look again at the diagram. That is the equation for dV, that is what you integrate to obtain the potential at P due to the line segment, its an integral over x.
     
  18. May 19, 2010 #17
    so for the ring r is constant
    so we dont treat it the same as that example of the line, we pull the r outside the intergral and intergrate?
     
  19. May 20, 2010 #18
    Correct, r is constant for the ring so you can pull it out of the integral. The integral is then just a line integral over the length of the ring which is just its perimeter.
     
  20. May 20, 2010 #19
    oh sweet
    so it would look somthing likkee
    krQ * (intergral(dr/2pir))
    with limits 2pi and 0?
    did i stuff up the Q? lambda = Q/l right? and Q is constant so i pulled it out
     
  21. May 21, 2010 #20
    would the angle between the dipole be 180 degrees or 90?
    if its 90 then its just 0
    but then if its 180 V will be negative
    and we dont know r

    V = k x (pcos(theta))/r^2)
    where p is the dipole moment
     
    Last edited: May 22, 2010
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