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Suppose the globe of a van de graff generator has a large positive charge. A small metal ball has a positive charge of 1.6 x 10 ^ -8 C, so the globe repels the ball. Assume that as average force of 6.0 x 10 ^ -3 N is needed to move the ball 8.0cm closer to the van de graff generator. How much work is done and what is the change in the electric potential of the ball?

I know how to do this:

W = Fd

W = (6.0 x 10 ^ -3)(0.08m)

W = 4.8 x 10 ^ -4 J

V = W/Q

= (4.8 x 10 ^ -4)/(1.6 x 10 ^ -8)

= 3.0 x 10 ^ 4V

Then there's a couple follow-up questions, Suppose the work done in moving the metal ball between the two points(8.0cm) is 6.5 x 10 ^ -4J, calculate the size of the repulsive force.

This I also did:

W = Fd

F = W/d

= 6.5 x 10 ^ -4/0.08

= 8.1 x 10 ^ -3 N

First off, I'd like to ask, the repulsive force of the van de graff generator must have a reaction force(newton's 3rd law), so then the small metal ball must also exert an equal reaction force on the van de graff generator, right? Is there also an action force exerted on the van de graff generator by small metal ball, and thus a raction force of the van de graff generator on the small metal ball? Or is the only action force the one the generator exerts on the ball?

Second, the second follow-up question asks the following:

suppose the change in electric potential of the charged ball remains 3.0 x 10 ^ 4V. Explain why the electrical potential should stay the same. What is the now the charge on the small metal ball?

I assume they mean to use the work mentioned in the previous problem to get the answer:

V = W/Q

Q = W/V

= (6.5 x 10 ^ -4)/(3.0 x 10 ^ -4)

= approximately the answer in the book, did I do that right?

Anyway, the 'explain' part is the confusing part for me. My teacher said it's because of coulomb's law: F(electrical) = (kQ_1Q_2)/(d ^ 2)

If you substitute that into the voltage equation you get the following:

V = W/Q

= Fd/Q

= (((kQ_1Q_2)/(d ^ 2)) x d)/Q_1

Q_1 cancels out, therefore proving that the charge on the small metal ball(Q_1) has no affect on the voltage. This could explain why the work is increasing, if Q_1 increases, force increases, therefore work increases, but voltage remains the same seeing as it cancels out in the voltage equation. Is this right?

There, a little long, but if someone could clarify my 2 questions(the last paragraph and the paragraph in the middle is where they are found) please? Please respond soon, thanks in advance.