1
$\begingroup$

When an electron moves from lower potential to higher potential and the work done by electric force is positive.

Therefore we will conclude that whenever a negative charge moves from a lower potential to higher potential, work done by electric force is positive and when a positive charge moves from lower potential to higher potential the work done by electric force is negative. Why? What is the concept?

$\endgroup$
  • $\begingroup$ Have you looked into the definition of "work"? $\endgroup$ – my2cts Apr 4 at 20:16
1
$\begingroup$

Therefore we will conclude that whenever a negative charge moves from a lower potential to higher potential, work done by electric force is positive

That is correct. The electron gains electrical potential but loses electrical potential energy. The energy for doing positive work on the electron comes from the electrical potential energy of the charge field system. If the electron is unrestrained it will accelerate and gain kinetic energy equal to the loss of electrical potential energy.

...and when a positive charge moves from lower potential to higher potential the work done by electric force is negative. Why? What is the concept?

Correct, but you are missing an important additional point.

An external (to the field) agent is needed exert a force to move a positive charge from lower potential to higher potential against the repulsive force of the electric field. Since the direction of the force exerted by the external agent is the same as the displacement of the charge, it does positive work. If the charge starts and ends at rest, the electric field does an equal amount of negative work since its force is in the opposite direction of the displacement. Assuming there are no dissipative forces involved, the net result is the negative work done by the field takes the energy supplied to the positive charge by the external agent and stores it as electrical potential energy.

The gravity analogy, in the absence of dissipative forces, is you do positive work lifting an object to rest at a height $h$ above the ground and gravity does an equal amount of negative work taking the energy you supplied and storing it as gravitational potential energy of $mgh$.

Hope this helps.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Well that's more clear 😁 $\endgroup$ – Hena Anvar Apr 5 at 7:28
  • $\begingroup$ Work done by external agents need not be stored as electrical potential energy. It can be dissipated as heat, as is done in resistors. The gravity analogy is incomplete if you are sliding the object up a wall. In that case you do more work to lift the object than is stored in the gravitational potential energy of the object. The rest is dissipated as heat due to the frictional force from the wall. $\endgroup$ – looksquirrel101 Apr 5 at 10:50
  • $\begingroup$ @lookssquirrel101 there are no dissipative forces involved in my example $\endgroup$ – Bob D Apr 5 at 11:31
  • $\begingroup$ @BobD - Yes, I am aware of that, which is why I stated that your analogy is incomplete. The point is that work done by an external agent is not necessarily stored as electrical potential energy, which is what you stated in the sentence prior to the analogy. $\endgroup$ – looksquirrel101 Apr 5 at 11:53
  • $\begingroup$ @looksquirrel101 Fine. I’ve included the assumption in my answer $\endgroup$ – Bob D Apr 5 at 12:07
2
$\begingroup$

The increment of work done by a force is $dW = {\bf F} \cdot d\bf x$.

The electric force on a charge is ${\bf F} = q {\bf E}$, where the electric field is ${\bf E} = -\nabla \phi$.

That means that the increment of work done on a charge by the electric force is $dW = -q\nabla \phi \cdot d{\bf x} = -q d\phi$.

So, by moving from a lower potential to a higher potential $d \phi$ is positive, and $-q d\phi$ is positive for a negative charge and negative for a positive charge.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Your answer assumes that the only force acting on the charge is the conservative electric force. What about friction forces? $\endgroup$ – Bob D Apr 5 at 12:27
  • $\begingroup$ @BobD No Bob, my answer makes no assumptions about any other forces acting on the charge. My answer provides the work done on the charge by the electrical force. That does not discount the possibility that there will be other forces on the charge that do work as well. $\endgroup$ – looksquirrel101 Apr 5 at 13:02
  • $\begingroup$ I respectfully disagree with you and leave it at that. $\endgroup$ – Bob D Apr 5 at 13:48
  • $\begingroup$ @BobD The math is irrefutable, so what is there to disagree with? The equations provide the work done by the electrical force on the charge. They have nothing to do with the work done on the charge by any other force that may exist. This is just basic mechanics. $\endgroup$ – looksquirrel101 Apr 5 at 18:07
1
$\begingroup$

Say, you've got a negative charge and a plate near it. Now suppose that the plate has a positive charge. So, you can say that the "location" of the positively charged plate is at a higher potential than the "location" of the negative charge.

Also, it is obvious that the positive plate will attract the electron towards it, and the electron would move from lower potential to higher potential i.e. the displacement is along the direction of electric force, and work done is positive

Now, in case you pull the electron farther away from the positive plate, there is going to be a decrease in its potential. The electric force is in the same direction (towards the positive plate) while the displacement is opposite and hence the work done is negative

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ how is the displacement along the electric field in the first case when it moves towards the plate it is from negative to positive or or is this talking about the electric line of force of a negative charge which is always radially inward $\endgroup$ – Hena Anvar Apr 4 at 18:24
  • $\begingroup$ The positive plate tries to pull the negative charge towards itself, hence the direction of force is towards the plate. In the first case, the displacement too is towards the positive plate. Hence, both are in the same direction $\endgroup$ – Krishna Apr 4 at 18:32
  • $\begingroup$ Electric field lines (or lines of force) of the negative charge do not cause any effect on the negative charge itself. $\endgroup$ – Krishna Apr 4 at 18:33
  • $\begingroup$ Okay thank you very much 😊 $\endgroup$ – Hena Anvar Apr 4 at 18:37

Not the answer you're looking for? Browse other questions tagged or ask your own question.