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It is given in the defination of electric potential in a website that we bring a charge from infinity to that point slowly(without acceleration). But if the charge placed at infinity is at rest or moving towards a random direction then how can we place the the charge on the required point without acceleration. May be I have some misconception , please clear my concept. I don't have much information so please help me out from this problem.

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An external force is applied to the charge which varies so as not to change the speed of the charge.
That external force can change the direction of motion of the charge but not change its speed by acting in a direction which is at right angles to the direction of the velocity of the charge.
This would mean that the force does no work on the charge and hence does not change its speed even though the charge is accelerating because it is changing its direction of motion.

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  • $\begingroup$ But what if charge placed at infinity is at rest.... $\endgroup$ – satyam bhardwaj Aug 31 at 8:58
  • $\begingroup$ It is clearly mentioned in its defination given by wikipedia that "without producing an acceleration ". $\endgroup$ – satyam bhardwaj Aug 31 at 9:04
  • $\begingroup$ Then how can we change the direction $\endgroup$ – satyam bhardwaj Aug 31 at 9:05
  • $\begingroup$ If the charge is at rest to start with the force can do some work to get the charge moving and the do the same amount of negative work to stop the charge at the final position. $\endgroup$ – Farcher Aug 31 at 10:48
  • $\begingroup$ I think that the no acceleration probably refers refer to not changing the speed (and hence the kinetic energy).of the charge. $\endgroup$ – Farcher Aug 31 at 10:49
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Obvious to move a charge from infinity without accelerating it is not a very practical proposal if the charges attract. You can also let it fall in, measure its speed and subtract the kinetic energy. For the repulsive case I see no issue other than that infinity is very far off.

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  • $\begingroup$ Can you please explain it more deeply....if you have a better link to study from then please send. $\endgroup$ – satyam bhardwaj Sep 1 at 15:07

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