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B Confusion about electric potential

  1. May 29, 2016 #1
    If I would displace a negative one-coulomb charge r meters from a positive charge, Q, the negative charge would gain electric potential energy that would be calculated by summing up all the values of the electric potentials along the line of its displacement, r. (I don't know calculus, but I have heard of integration.) Why then is the formula for electric potential KQ/r? Doesn't this come from multiplying the electric force at displacement r from Q with r? Why do we do that when the electric force is actually changing along the path of displacement and not constant? Is not this, for example, like multiplying the final acceleration of an object by the duration of the acceleration to find the final velocity, which is invalid because the acceleration is actually constantly changing along the path?
    Last edited: May 29, 2016
  2. jcsd
  3. May 29, 2016 #2


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    Staff: Mentor

    If you actually calculate out the integration, you'll find that if the force at every point goes as ##1/r^2## the potential at any point goes as ##1/r##.

    The change in potential energy is the difference between the potential at the starting point and the potential at the ending point. It's also what you get by summing (as an integral) the change in the potential across each infinitesimal step along the path from starting point to ending point.

    Do not confuse the change in the potential between two points separated by a distance ##r## and the potential at a point at a distance ##r## from the positive charge. The best way to keep it all straight is to use different letters for the different quantities in the problem:
    ##R_0##: the distance of the starting point from the central charge.
    ##R_1##: the distance of the ending point from the central charge
    ##\Delta{R}=R_1-R_0##: the distance the negative charge is displaced
    ##r##: a label for an arbitrary distance from the central charge. The potential energy at a point at a distance ##r## from the central charge is ##-KQ/r## and the attractive force is ##KQ/r^2##.

    The potential energy gain when I move the charge a distance ##\Delta{R}## from the starting point at ##R_0## to the ending point at ##R_1## is ##KQ(1/R_1-1/R_0##.
  4. May 29, 2016 #3
    So should I take this for granted till I study calculus? Is not there any other way to understand it without calculus?
  5. May 29, 2016 #4


    Staff: Mentor

    I don't know of another way. I suppose that you could graphically approximate it to convince yourself that it is correct.
  6. May 31, 2016 #5


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    Science Advisor
    Gold Member
    2017 Award

    Physics without calculus is much more difficult than physics with calculus. That's why Newton succeeded over all his predecessors and created a new kind of science, i.e., (theoretical) physics. So it's well spent time to learn calculus first!
  7. Jun 1, 2016 #6
    You should not be too concerned just yet about lack of calculus knowledge. The Physics ideas are straightforward.
    You realise that point charges produce FORCES that are proportional to 1/r2 (inverse square law). POTENTIAL (energy) is a measure of work done and is therefore indicated by the area under F ~ distance graph. It is not difficult to verify that the area under 1/r2 graphs is proportional to 1/r.
    This can be done manually by 'counting squares' under the 1/r2 graph. This is also known as integration (dreaded calculus) and for common functions such as 1/r2 standard formulae are well known.
    Understanding calculus is undoubtedly a great 'tool' in the understanding of physical principles but the principles are not calculus !
    Another majot aspect to be aware of is that the zero of POTENTIAL energy is taken to be at infinity (true also in gravitational fields) and this means that in some cases (such as gravitation) increasing POTENTIAL can give a max value of zero. In this case POTENTIAL must be a negative quantity
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