The work done by an external force to move a -6.50muC charge from point A to point B is $15.0*10^{-4}J$, if the charge had started from rest had $4.82*10^{-4}J$ of kinetic energy when it reached point B, what must be the potential Difference between A and B?

Dont we just need the Equation $qVba = -W_{electric}$? since it is asking for the potential difference, why would we need the kinetic energy at point B? We are not calculating the potential at B or A anyways. Can someone enlighten me on what I'm thinking wrong?

  • $\begingroup$ I've added the homework-and-exercises tag. In the future, please use this tag on this type of question. $\endgroup$
    – user4552
    Feb 27, 2019 at 15:02
  • $\begingroup$ Since energy is conserved, the kinetic energy possessed by the charge at point B is due to the conversion of some of its potential energy to the kinetic one. And the reason why the charge can convert its energy potential is because the two points (A and B) have a potential difference. So collecting these pieces of information you should be able to derive an equation to get the potential difference. $\endgroup$
    – rnels12
    Feb 28, 2019 at 11:41


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