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Homework Help: Electric potential - plot,

  1. Jun 5, 2008 #1
    what i really want to ask is, what is the problem asking:

    - do o find the potential difference between for each charge? which pairs?

    - do i plot the potentials on a graph?

    -what do they mean "as a function of x along the x-axis"? are they talking about the x-axis of the graph i'm supposed to plot?


    plot the electric potential as a function of x along the x-axis for the following configuration of electric charges --> microC = microcoulomb, * = electric charge

    +2 microC -3 microC +1 microC
    x = 0cm x = 4cm x = 6cm

    2. Relevant equations

    potential difference V_ab = kq ( (1/r_b) - (1/r_b) ) where k = 9*10^9 constant, q = charge, Coulombs, r = radius in meters

    3. The attempt at a solution

    can't really make an attempt until i know what the question is asking, do i just find the electric potential differences between the various charges above, using the above eq and plot them such that potential V is on the y-axis, and charge is on the x-axis?

    help appreciated...
  2. jcsd
  3. Jun 6, 2008 #2


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    The question is asking for the value of the electric potential at each point along the x-axis. You have three electric charges, so you are being asked to find the sum of the individual potentials of the charges. (The nice thing about doing this for electric potential is that you just have to add up numbers, rather than vectors.) So you don't want potential differences; you just want to calculate kq/r for each charge at any value of x and add up the values. (You will, of course, have infinities at the positions of the charges themselves...)
  4. Jun 6, 2008 #3
    thanks, but what do i use for distance r? i'm assuming just using the given distances for each charge, i am not finding the potential between two charges, just one correct?

    using the eq specified V = kq/r

    for the charge at x = 0cm, i got a potential of 0
    for the charge at x = 4cm, i got a potential of -675000
    for the charge at x = 6cm, i got a potential of +150000

    so for the sum i got -525000, what were the units again? volts/meter?

    is my result logical? now do i have to plot a graph using this data? what do they mean by 'plot as a function'?

    Last edited: Jun 6, 2008
  5. Jun 6, 2008 #4


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    As I read your diagram, there are charges at x = 0, x = 4 cm, and x = 6 cm. The distances you use in your electric potential calculations are the distances of a chosen point x from each of these three positions.

    The units for potential, BTW, are volts, since k has the units N·(m^2)/(C^2) = J·m/(C^2) = V·m/C .

    If we want to find the total potential at, say, x = +1 cm, the distances from each charge will be 1 cm. = 0.01 m. from the charge at x = 0, 3 cm. = 0.03 m. from the charge at x = 4 cm., and 5 cm. = 0.05 m. from the charge at x = 6 cm. So the individual electric potentials at x = +0.01 m. will be

    (9·10^9)·(2·10^-6)/0.01 m from the charge at x = 0 cm. ,
    (9·10^9)·(-3·10^-6)/0.03 m from the charge at x = 4 cm. , and
    (9·10^9)·(1·10^-6)/0.05 m from the charge at x = 6 cm.

    You would then add up these values, one of which is negative, to find the total electric potential at x = +1 cm. You would do something similar for any other position x on the line. What you would plot would be the value of the total potential V (in volts) as a function of the position x. (You could also set up a set of terms for this and use it on a graphing program.)

    At the position x = 0 , the distance from the charge at x = 0 is zero, so the potential would be (9·10^9)·(2·10^-6)/0 , which gives positive infinity. So your total potential will have "spikes" in it at x = 0, 4, and 6 cm., which each charge is located.
  6. Jun 6, 2008 #5
    i understand what you are saying about potential in terms of at say x = +1cm, but if i am correct, the problem does not ask me to find potential difference between two locations, but just the potential at each of the three given points.

    on the graph, what do i label the axis, electric potential of each charge on the y-axis, and distance from the origin on the x-axis?

    i still don't understand what they mean by 'as a function,' could you please clarify?

    Last edited: Jun 6, 2008
  7. Jun 6, 2008 #6


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    This is not calculating potential differences between points; we are calculating the value of total electric potential at a point x along the line which is not at the location of any of the charges. Each charge contributes a potential V = kq/r at a point x, given by

    V_1 = k·q_1 / (x - 0) Volts,
    V_2 = k·q_2 / (x - 0.04) Volts, and
    V_3 = k·q_3 / (x - 0.06) Volts,

    where x is the location of a point on the line, in meters.

    The total electric potential is then V = V_1 + V_2 + V_3 . Since this sum depends on the value of x for a point we choose on the line, we say that V is a function of x.

    What you are being asked to plot is that sum V against a position on the x-axis, represented by the coordinate x. The total electric potential will have positive and negative values (and be zero at some points), and will run away to positive or negative infinity at the location of the charges.

    A potential difference would be the difference in electric potential between two locations on the x-axis. So, for instance, the potential difference between the point at x = -10 cm. and x = +10 cm. would be V(-0.1 m) - V(+0.1 m). You are not asked to plot that, but you could figure it out easily from your graph of V(x).

    You are also not being asked to find the potential at the location of the charges: those would be positive or negative infinity.
  8. Jun 6, 2008 #7


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    The value of the potential is different for each value of x. Given x, it is possible to calculate the potential at x. We describe such a situation with the words: "the potential is a function of x".

    They just want a graph with potential on the y-axis, and x on the x-axis. You've been shown how to calculate one point on the graph, the potential at x = 1 cm.
  9. Jun 6, 2008 #8
    now i'm confused but i do understand the whole 'function' thing now, thanks. if i use the equations:

    V_1 = k·q_1 / (x - 0) Volts, ------------>where x =0m
    V_2 = k·q_2 / (x - 0.04) Volts, and --------->where x = 0.04m
    V_3 = k·q_3 / (x - 0.06) Volts, ---------------> where x = 0.06m

    wouldn't i just a get a net V of zero since the denominators for each one will be zero? then the graph would just be a of nothing.

    am i seeing something incorrectly?

    what i did originally was this:

    V_1 = kq_1/x = 0 where x = 0m
    V_2 = kq_2/x = -675000 where x = 0.04m
    V_3 = kq_3/x = +150000 where x = 0.06m

    so V_sum = -525000

    these numbers are much more 'graphable.' what now?.....
  10. Jun 6, 2008 #9


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    You do not put the value of x for each charge into the terms for the potentials; rather, you choose a particular position on the
    x-axis and use its coordinate value to calculate each of the potentials. For x = + 1 cm = +0.01 m., this will give

    V_1 = (9·10^9)·(2·10^-6)/|0.01 - 0| = +1,800,000 V ,
    V_2 = (9·10^9)·(-3·10^-6)/|0.01 - 0.04| = -900,000 V , and
    V_3 = (9·10^9)·(1·10^-6)/|0.01 - 0.06| = +180,000 V .

    This makes the total electric potential at x = +0.01 m. equal to the sum,
    V(+0.01 m.) = +1,080,000 V . (And I realized that the general expression should have absolute value signs, since we want the distance of the point x from each charge with no regard to sign.) You would do a similar calculation for every value of x.

    As for your examples, you don't want to use x = 0, 0.04 m, or
    0.06 m. The first of these will give you

    for x = 0 m., V_1 = k·q_1 / (0 - 0) Volts = k·q_1/0 = +infinity!

    By the same token, at x = 0.04 m., you'd have

    V_2 = k·q_2 / (0.04 - 0.04) Volts = k·q_2/0 = -infinity ,
    since q_2 < 0 .

    This makes the total electric potential undefined at the locations of the charges, so we do not do the calculation at those places.

    The point is that you use a single value for x in all three charge potential terms and add those values together to get a total, not separate values for each charge.
    Last edited: Jun 6, 2008
  11. Jun 6, 2008 #10
    oh thanks that makes more sense, so basically i choose arbitrary x values, other than x = 0, 0.04, and 0.06m, correct?

    so say i first choose x = 0.01m, i then sum them up using the three general equations and plot the sum and x=0.01, then do i choose another arbitary x value and do the same thing, until a usable plot develops, yes?

    the number of x values i choose is up to me, i'm just supposed to choose enough x values so that i have enough data to plot something usable, correct?

  12. Jun 6, 2008 #11


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    Yes, just so!

    Keep in mind that the denominators of the potential terms use the absolute values of distance, because it doesn't matter for electric potential whether the location of the "test point" is to the left or right of the charge. You can even use something like Excel to calculate the values, since the algorithm is well-defined and you need to find many results to make a plot. If you use something which cannot give absolute values, there is a trick to "fool" the software (I had to use it on the graphing program I have). You write each potential term as

    k · q / [sqrt{ (x - 0.04)^2 }] , for instance,

    the idea being that the software always finds the positive square root and squaring the distance first eliminates the sign of the (x - 0.04) term, so the square root gives the absolute value |x - 0.04| .

    You will find, I believe, that your potential function is zero at three positions. You may also want to plot the potential V in millions of volts (divide your calculated results by 10^6), because the values can get quite large (indeed, they will race off to infinity when you're close to any of the charges).