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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 $F_{me}$ , $F_R$.

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  • $\begingroup$ 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. $\endgroup$ – The Photon Aug 12 at 1:46
  • $\begingroup$ Correct, for the inertial change. Thanks for pointing that out! $\endgroup$ – mai Aug 12 at 14:03
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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 $F_{q2}$.

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  • $\begingroup$ Interesting that $F_{q_2}$ = $F_{mc}$ could you explain that? $\endgroup$ – mai Aug 12 at 14:04
  • $\begingroup$ I assumed them as two separate vectors to contribute to $F$. $\endgroup$ – mai Aug 12 at 14:05
  • $\begingroup$ 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. $\endgroup$ – Pox 219 Aug 12 at 14:12
  • $\begingroup$ Thanks for the clarification and answer. $\endgroup$ – mai Aug 13 at 14:04
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BEST WAY TO understand it by work energy theorm

what work energy theorm state-The work W done by the net force on a particle equals the change in the particle’s kinetic energy KE: so there are two forces that is working first Fm,second electrostatic now the net work done will be equal to to sum of both that are opposite direction net work done = 1/2 mv2(as you described that you move rapidally) if work done by both forces equal then K.E=0 but it is not that mean Fme should > FR and F = Fme - Fq2=ma

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