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In a vacuum, a uniform electric field of strength $E$ is applied in the positive x-axis direction. When you carry a charged particle with a positive charge $q$ from $A$ to $B$, seek the work that the electrostatic force did.

My textbook says the external force and the electrostatic force have the same magnitude.

What is the External force?

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closed as off-topic by John Rennie, Jon Custer, tpg2114 Oct 2 at 21:13

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What your book is saying, for moving the block from B to A you need to do some work, and this work will be same and opposite of the work done by electrostatic force. Here external force, which provide by you and internal mean electrostatic.

You can think about this situation in one more way, if charge to be at rest, at every point on line Ab=force by external agent should be equal to force by electric field.

Edit:I agree with Bob

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  • $\begingroup$ If you want I can go through the calculation. $\endgroup$ – yuvraj singh Oct 2 at 9:13
  • $\begingroup$ Thank you for your reply. Actually, my question is why the external force should be equal to the electrostatic force. If they are not the same, what will happen? $\endgroup$ – MENZIES Oct 2 at 9:30
  • $\begingroup$ Now let me clear my answer, why I say Fext=Felectrostatic, if this this is not the case, mean fext>Felectrostatic there will be additional k. E to charge, and you will not get true potential energy at point B . then we have to do more calculation $\endgroup$ – yuvraj singh Oct 2 at 9:34
  • $\begingroup$ I see. This question does not say there is no change of potential energy, but we know that there is no change because there are unspoken rules. Is that right? $\endgroup$ – MENZIES Oct 2 at 9:40
  • $\begingroup$ There will be change in potential energy when you from A to B. $\endgroup$ – yuvraj singh Oct 2 at 9:51
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I think there is something wrong with the problem. By convention the direction of the electric field is the direction of the force that a positive charge would experience if placed in the field. In this case no external force is needed to move the charge from A to B as it would naturally be accelerated by the field. Only the field does work.

If the charge had to be moved from B to A, an external force would be needed to move the charge against the repulsive force of the field. That requires positive work by an external force. At the same time the field does negative work because its force is in the opposite direction of the movement of the charge, taking the energy supplied by the external force and storing it as electrostatic potential energy. Then Yuvraj answer would apply.

UPDATE:

The above said, there is another explanation for what the problem is getting at. If the charge is simply placed at A it will accelerate to B losing potential energy and gaining kinetic energy, as said above. But if at the same time an external force is applied to the charge equal to and opposite the force of the field bringing the charge to rest at point B, the charge will have moved at constant velocity from A to B and the overall change in kinetic energy is zero. Now the field does positive work and the external force does an equal amount of negative work taking the potential energy away from the charge.

Gravity analogy:

The two scenarios have a gravity analogies.

In the first scenario, the charge moving from A to B under the influence of the field without opposition is analogous to the work done by gravity on a free falling object. The gravitational field does net positive work giving the object kinetic energy at the expense of gravitational potential energy.

The second scenario is analogous to taking an object at rest a height $h$ above the ground and slowly lowering it by applying an upward force equal to $mg$, bringing the object to rest on the ground, instead of free fall. The gravitational field does positive work on the object and my external force does an equal amount of negative work taking the gravitational potential energy away from the object.

Hope this helps

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  • $\begingroup$ Thank you for your advice. What if there should not be additional K.E., there should be the negative external force. Now, I think that my textbook want to say this. What do you think about this? $\endgroup$ – MENZIES Oct 2 at 11:14
  • $\begingroup$ Yes. The key is the magnitude of the external force has to equal the magnitude of the force of the electric field such that the charge moves at constant velocity from B to A. Then there is no change in kinetic energy going from B to A which means the net work done is zero per the work energy theorem. The thing is, this all applies if the charge is moved from B to A, not A to B, which is why I have an issue with the problem statement. Yuvraj's latest edit agrees with that point. $\endgroup$ – Bob D Oct 2 at 11:53
  • $\begingroup$ @MENZIES There is another possible explanation for the problem. I will revise my answer to include it. $\endgroup$ – Bob D Oct 2 at 13:31
  • $\begingroup$ Thanks. Your explanation helps me very much. $\endgroup$ – MENZIES Oct 2 at 14:00

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