For this exercise, imagine a large positive charge +Q fixed at some location in

Question

For this exercise, imagine a large positive charge +Q fixed at some location in otherwise empty space, far from all other charges. A positive test charge of smaller magnitude +q is launched directly towards the fixed charge. Of course, as the test charge gets closer, the repulsive force exerted on it by the fixed charge slows it down. Your job is to explain why the test charge slows down, but in terms of electric potential and EPE, rather than in terms of fields or forces.

In particular:
• explain how the electrical potential encountered by the test charge changes as it gets closer to the fixed charge, and why
• explain how the EPE of the test charge changes as it gets closer to the fixed charge, and why

On that basis, explain why the test charge +q slows down as it approaches the fixed charge +Q.

Note
Your response shouldn't mention either fields or forces.
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THIS IS THE ANSWER I TURNED IN:
“As the test charge is projected towards large positive charge Q with full kinetic energy, its potential energy at the begining is zero.
Total mechanical energy=KE+PE
at the begining total energy is only KE.
Law of conservation energy states that total energy of the object remains constant as the object moves, provided that net work done by external nonconservative force is zero.
PE increse because work done against repulsion (conservative force) is increasing as test charge approaches fixed large positive charge.
To balance total energy, KE slowly decrease. so the test +q is is slowing down.
On reaching a minimum distance total energy is only PE.
EPE=kQq / r where r is minimu distance of test charge (+q) from fixed large charge(Q)”

THIS WAS THE INSTRUCTORS RESPONSE:
This is pretty good! You've overall got the right idea. What's missing is a discussion of how the test charge's EPE is related to the change in potential (V) it experiences as it moves closer to the fixed charge. You're discussing this in terms of work and forces, but forces have been ruled out.

Explanation / Answer

The only things that I would add to your response: "electric potential" V was devised because it is a convenient way of expressing potential energy per unit of charge. You have your large fixed positive charge +Q; once you know the potential at a point around it, all you have to do is multiply that potential by whatever small charge q you place at that point in order to calculate the potential energy of the two-charge system. If you place a incoming q ten times larger than another -- you have ten times as much potential energy. The other point you might want to make is that a basic law of nature is that systems seek low energy. Everything runs downhill. And that's why your small charge doesn't want to get too close to the large charge. The potential gradient translates into a potential energy gradient which makes the route in to the big charge uphill all the way (height increasing with 1/r). Things don't like to go uphill.

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