My revision notes states that "to move the positive test charge from infinity to point P, the external force provided to that test charge must be equal to the electric force the source charge exerts on the test charge". In the case of a positive source charge, I'm not sure whether the quote above means:

  1. A constant external force is exerted on the test charge, which is more than the electric force at infinity, but becomes equivalent to the electric force at a point closer to the source charge, r; or,

  2. The external force increases with the electric force exerted on the test charge as it is moved from infinity to r.

Option 1 seems most likely. Since at infinity, it is likely the test charge is at rest since it is outside the electric field of the positive source charge. According to Newton's first law, the external force that is exerted on the test charge needs to be more than the "electric force" exerted on that charge in order to result in it moving.

However, my notes also state that "there is no change in K.E., hence there is no acceleration." If option 1 is the case, then there is a change in K.E., except the K.E. decreases since the acceleration decreases (resultant force decreases as electric force increases) as the charge is moved from infinity to the point, r.

For option 2, there is no change in K.E. since the charge is in constant velocity, but initially at infinity, according to Newton's first law, if the external force is the same as the electric force, the resultant force is zero which means the charge stays at rest (assuming that the charge is stationary at infinity). So I'm not quite sure which option is correct.

Also, why must K.E. be kept constant? Technically, the work that is being done is constant which should contribute to both the potential energy and kinetic energy of the test charge?

  • $\begingroup$ The answer is given here for a gravitational field. physics.stackexchange.com/a/382385/104696 You either have the test charge moving are constant velocity or start with a stationary test charge at the start, give the test charge some kinetic energy so the it can undergo its journey and just before the end take away all that kinetic energy so that the test charge finishes up at rest when it is at its destination. $\endgroup$
    – Farcher
    Jun 10 '18 at 5:01
  • $\begingroup$ why must K.E. be kept constant? $\endgroup$
    – xander
    Jun 10 '18 at 9:05
  • $\begingroup$ Keep $v$ constant, so that there's no acceleration, which means all the force is going to fight the electric field, and no portion of force is going to accelerate the charge. $\endgroup$
    – FGSUZ
    Jun 10 '18 at 9:12
  • $\begingroup$ It is so that all the work done by the external force is related to moving the test charge in the electric field. $\endgroup$
    – Farcher
    Jun 10 '18 at 9:12

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