I was watching a video about Electrostatic Potential and Electric Potential Energy by Professor Walter Lewin when I encountered a question.
So, Professor Lewin assumes that a charge $Q$ was placed in free space, and hence, no work was done to place it there. Then, another charge (a test charge) $q$ was placed at a distance of $R$ from $Q$. Now, some work would have to be done by the external agent placing $q$ near $Q$.
We say that the charge was brought from a distance of $\infty$ from $Q$ to a distance of $R$ from it.
The work that would be done by the external agent to do so would be equal to the potential energy possessed by $q$ at a distance of $R$ from $Q$.
So, we just need to evaluate the work done in doing so.

Professor Walter Lewin uses the formula for calculating work, specifically : $$W = \int_a^b \vec{F \text{ }} . \vec{dr}$$ Here, $a = \infty$ and $b = R$. Also, $\vec{F}$ is the force exerted by the external agent. Let that force be $\vec{F_2}$ in the direction opposite to that of $Fe3$ Professor Walter Lewin proceeds as follows : $$\int_\infty^R \vec{F_2} . \vec{dr}$$ Now, he states that $\color{red}{\text{the force exerted by the external agent is equal in magnitude but opposite in direction to the electrostatic force}}$, that is $F_e$. So, $\vec{F_2} = -\vec{F_e}$. He proceeds as follows : $$\int_\infty^R \vec{F_2}.\vec{dr} = \int_R^\infty \vec{F_e}.\vec{dr} = \int_R^\infty \dfrac{Qq}{4\pi\varepsilon_0}.\vec{dr}$$ He simplifies the integral obtained above as $\dfrac{Qq}{4\pi\varepsilon_0R}$

Now, I was confused by the statement that I have highlighted in $\color{red}{Red}$

If the forces acting on the charge $q$ are equal in magnitude but opposite in direction at all points from $R$ to $\infty$, doesn't that cancel them out? Doesn't that mean that there will be no displacement at all and hence, no work done.
What I understood the force $\vec{F_2}$ prior to watching this video was as a force which increases as the distance between $Q$ and $q$ approaches $R$ so as to always be greater than $\vec{F_e}$ at every point by a constant amount. But that confused me, hence, I came to Professor Walter Lewin's video.

I would appreciate it if someone can clarify the relationship between $\vec{F_e}$ and $\vec{F_2}$ for me and help me to get rid of my misconceptions.

  • $\begingroup$ There is no change in Kinetic energy of the charge particle. $\endgroup$ Jun 29, 2020 at 5:28
  • $\begingroup$ @Ohw Can you please elaborate a little? Thanks! $\endgroup$ Jun 29, 2020 at 5:29

2 Answers 2


Yes that's how I have been taught about it.

But what it actually refers to is that the object is not allowed to accelerate and hence not allowed to gain velocity during its movement.

This is because in the case that it's allowed to gain velocity, it has a significant kinetic energy. Which should also be factored in. You could also proceed this way though it is unorthodox

The only way to avoid factoring in the kinetic energy is by making it move such that it doesn't gain velocity.

So if a force $F$ (electrostatic) exists, then external force should be $F+dF$ in the opposite direction.

This $dF$ force allows it to move slightly, but

$$dW = (F+dF). dx = (F. dx) + (dF. dx)$$

The $dF. dx$ term is so small that it is negligible (As done in several instances in calculus)

So you get the answer mentioned in the video.

  • $\begingroup$ Thank You for the answer, first of all. I do have a question, hope you don't mind. Is the reason to why $\int^R_\infty \vec{F_2} .\vec{dr} = \int_R^\infty \vec{F_e}.\vec{dr}$ because the sum of the little work(s) $(dW)$ done by $F_2$ going from $\infty$ to $R$ will be the same as the sum of the little work(s) done by $F_e$ in going from $R$ to $\infty$ as the magnitude of $F_e$ is approximately equal to that of $F_2$. I know this may sound obvious but I want to make sure that I have no misconceptions left before I move on. Thanks! $\endgroup$ Jul 3, 2020 at 4:01
  • $\begingroup$ @RajdeepSindhu If you want a realistic case, you can move it randomly, stop it and do anything you can with an external force. But when you reach the desired point, you stop. This ensures that final kinetic energy is 0. And for your question, yes we take an approximate value there, however it is not required if you just ensure the object "stops " finally. This nullifies the approximation $\endgroup$ Jul 3, 2020 at 5:03

Work Energy theorem tells us work done by all forces is equal to change in kinetic energy of the particle.

Now we bring the particle so slowly((without accelerating)) that there is no change in kinetic energy of the particle i.e $\text{change in kinetic energy of the particle} = 0$

Hence, $\text{work done by electrostatic force} + \text{work done by external force} = 0$


$|\text{work done by electrostatic force}| = |\text{ work done by external force}|$. Now, since displacement is same we can say that those forces are equal but opposite in direction.

  • $\begingroup$ So, basically, $|\vec{F_e}| \approx |\vec{F_2}|$? $\endgroup$ Jun 29, 2020 at 5:37
  • $\begingroup$ Not approximately but exactly otherwise charge particle will experience net force , will accelerate and thus will result in change in KE. $\endgroup$ Jun 29, 2020 at 5:38
  • $\begingroup$ No ! there is no change in KE at all. It's the way we defined electrostatic potential. $\endgroup$ Jun 29, 2020 at 5:41
  • $\begingroup$ I think I'm missing something. If the work done by external force is $0$, doesn't that imply that the electrostatic potential energy is $0$ since electrostatic potential energy at a distance of $R$ from a charge is defined as the work done by an external agent in bringing the test charge from a distance of infinity to R against the electrostatic force exerted by the positive charge on the test charge? $\endgroup$ Jun 29, 2020 at 5:46
  • 1
    $\begingroup$ @RajdeepSindhu Sorry, it was a typo. $\endgroup$ Jun 29, 2020 at 5:56

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