# Accurately calculating the work to move a charge? [on hold]

I'm applying a force $$F_{me}$$ to move the charge $$q_2$$ towards $$q_1$$.

If they are both equal charges, I'm assuming a repulsive force $$F_R$$ opposing me.

In order to get $$q_2$$ to the final position required($$P$$), I believe the work equation would be as such:

$$W_{me} = \int_{r} F \cdot dr$$

What if I needed to move $$q_2$$ rapidly to $$P$$, wouldn't $$F$$ be:

1. $$F_{me}$$ > $$F_R$$
2. $$F_{q_2}$$ = $$m_{q_2} a_{q_2}$$

$$F$$ = $$F_{me}$$ + $$F_{q_2}$$

Applying coulomb's law for $$F_{me}$$ , $$F_R$$.

## put on hold as off-topic by Aaron Stevens, Jon Custer, ZeroTheHero, Gert, Kyle Kanos18 hours ago

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, Jon Custer, ZeroTheHero, Gert, Kyle Kanos
If this question can be reworded to fit the rules in the help center, please edit the question.

• If you want to stop the charge when it gets to P, the extra component of F will have to point back the other way. – The Photon Aug 12 at 1:46
• Correct, for the inertial change. Thanks for pointing that out! – mai Aug 12 at 14:03

I believe that $$F= F_{q2} = F_{mc} - F_{R}$$, as the force in Newton's 2nd law is the total force applied to an object. In this case, the forces applied to the particle are $$F_{mc}$$ and $$F_{R}$$, resulting in $$F_{q2}$$.
• Interesting that $F_{q_2}$ = $F_{mc}$ could you explain that? – mai Aug 12 at 14:04
• I assumed them as two separate vectors to contribute to $F$. – mai Aug 12 at 14:05
• I didn't say $F_{q2} = F$. The way I thought about my answer it is as follows. There is a repulsive force, $F_R$, and a push force, $F_{mc}$. Just like when we resolve forces for, say, a falling object, the sum of ALL the forces on the particle = $F_{mc} - F_R$ = $ma$. Newton's 2nd law is only applicable once you've summed up all forces in a direction. – Pox 219 Aug 12 at 14:12