While studying electric potential, I run into an issue on exercise 43 of Chapter 24 of Fundamentals of Physics 8th ed. Vol. 3 (Halliday et al.).

The exercise states the following, paraphrased: A charged particle (q = 7.5uC) is released from rest at point 60cm on the x axis. It moves 40cm due to another charged particle (Q) at point 0cm (origin) on the x axis. The exercise asks what is the kinetic energy of the particle (q) after the displacement, considering: (a) Q = 20uC and (b) Q = -20uC.

To solve the exercise I used the following: $E_{kf} = -q * \Delta V$

I arrived at the intended results: (a) Ecf = 0.9 joule and (b) Ecf = 4.5 joule.

My question is: Is it possible to reach the results by using the electrical work formula $W_e = F_e * ds$ instead? I've tried using it, but I do not get the same results. Does it require some data not informed by the exercise? Is it applicable to this situation? Am I misinterpreting something?

Thanks in advance.


1 Answer 1


You should be able to get the same result by using the electrical work formula - but note that you need to integrate since the force changes with position. That's really all the potential is - it is the integral of force for unit charge. That's why force has the $1/r^2$ relationship while potential has $1/r$ (with appropriate signs and constants...).

Perhaps this is enough. If not, then show how you tried to use the electrical force equation to solve the problem.

  • $\begingroup$ Thanks very much, I was calculating the wrong way. But now that I integrate it, the results are the same. $\endgroup$
    – Ike
    Oct 29, 2014 at 0:09

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