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I am teaching myself Electrodynamics through Griffiths' text, but am having trouble understanding how much work it takes to move a test charge $Q$ from a point $\mathbf{a}$ to a point $\mathbf{b}$ in the presence of a stationary configuration of source charges.

So suppose that we have a stationary configuration of source charges, and that we would like to move a test charge $Q$ from $\mathbf{a}$ to $\mathbf{b}$ along a path $C$. Let us first suppose that $\mathbf{a}$ and $\mathbf{b}$ are finite points in $\mathbb{R}^3$. We know that the source charge configuration generates an electric field $\mathbf{E}$, and so the force on the test charge $Q$ by the source charges is given by $\mathbf{F}=Q\mathbf{E}$. Griffiths states that the work required to do this is $$W=\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{F}\cdot d\mathbf{l}=-Q\int_{\mathbf{a}}^{\mathbf{b}}\mathbf{E}\cdot d\mathbf{l}.$$

I understand the negative sign is there because we are opposing the force generated by the source charge configuration, but I cannot understand how the particle can accelerate if it starts from rest. For if the particle starts at rest and we oppose the with an equal an opposite force, then $$F_{\mathrm{net}}=Q\mathbf{E}-Q\mathbf{E}=\mathbf{0},$$ and there is no acceleration.

How can we bring the particle to motion from rest, while applying a force that is only equal and opposite. It seems that for some time, maybe an infinitesimal time, the force would need to be stronger. And if that's true how can work be given by the formula above?

Furthermore, suppose that we "bring the charge in from infinity". Does this problem go away? It seems like it would, because then the particle could at least have a constant velocity. So the forces could sum to zero, but we could still move the particle from infinity to $\mathbf{b}$ without acceleration.

I have been stumped on this for quite a while, and would be very grateful for anyone who could provide some help.

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You are right, but it is implicit in the phrase "work required to move a test charge $Q$ from $\mathbf{a}$ to $\mathbf{b}$ along a path $C$" that this is the ideal minimun work required, that is, if you can neglect the work that you have to do to accelerate the particle from rest ($W_{\text{extra}}\approx 0$).

For example, if the charged particle is initially at rest, you can obtain that $$W=-Q\int_{C}\mathbf{E}\cdot\text{d}\mathbf{l}+W_{\text{extra}}\approx-Q\int_{C}\mathbf{E}\cdot\text{d}\mathbf{l}$$ if you move the particle extremely slowly, so that the extra work $W_{\text{extra}}$ that you have to do to start the movement and to change the direction of the particle in order to follow the path $C$ is negligible (one way to accomplish this is by satisfying the condition $\mathbf{F}_{\text{net}}\approx Q\mathbf{E}-Q\mathbf{E}=\mathbf{0}$ in each point of $C$).

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  • $\begingroup$ Thanks for your very helpful answer! But why would you need extra direction to change the direction of the particle? Once it is going, why can't the force $-Q\mathbf{E}$ propel it in the necessary direction? Doesn't it have to, since it is equal and opposite to $Q\mathbf{E}$? $\endgroup$ Jun 15 '20 at 17:54
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    $\begingroup$ I think that you are asking about the $W_{\text{extra}}$ that you need in order to change the direction of the particle. That extra work would come from the fact that you want the charged particle to follow the path $C$ and not the path given only by the initial conditions of the particle and its interaction with the electric field. This implies that you must do work not only to start the movement and to oppose to the force due to the electric field, but also to make the particle follow the path $C$. $\endgroup$
    – NarcosisGF
    Jun 15 '20 at 18:16
  • $\begingroup$ But once that initial work to get the particle on the path of $C$ is done, no additional work is required? Am I understanding this correct? $\endgroup$ Jun 15 '20 at 18:32
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Like you said, if you apply a force perfectly opposed to the force of the electric field on your particle, the net force will be zero, meaning your particle will either stand still or move in a straight line at constant velocity. The latter case would be fine if your path $C$ is a straight line from $\mathbf a$ to $\mathbf b$.

If your particle starts from rest at point $\mathbf a$, like you said, it would need a little kick to get it going. Consider a brief force that does positive work $W_0$, giving your particle some velocity. Here's the key: at point $\mathbf b$, the velocity of your particle should also be zero. In order to stop your particle, you'll need to decrease its velocity by the same amount you just increased it, which would require a negative work precisely equal to $W_0$. Essentially, all the extra work that actually moves your particle from $\mathbf a$ to $\mathbf b$ comes out in the wash.

The interesting thing is that this is true regardless of how your particle gets from $\mathbf a$ to $\mathbf b$. There's actually no reason your particle needs to have a constant velocity along its path. So long as there's no loss of energy, whatever forces move your particle along its path will always cancel in such a way that they do precisely the amount of work required to oppose the electric field (so long as its kinetic energy is the same at both ends).

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  • $\begingroup$ Thank you for your very helpful answer! But I'm not sure why "There's actually no reason your particle needs to have a constant velocity along its path". If all the forces sum to zero, how can there be any acceleration? And I am not sure what you mean by "So long as there's no loss of energy, whatever forces move your particle along its path will always cancel in such a way that they do precisely the amount of work required to oppose the electric field". $\endgroup$ Jun 15 '20 at 17:50

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