Question about electrostatic forces

[breedb]

Two identical conducting spheres, fixed in place, attract each other with an electrostatic force of 0.0864 N when their center-to-center separation is 70.6 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic force of 0.0422 N.



Of the initial charges on the spheres, with a positive net charge, what was (a) the negative charge in coulombs on one of them and (b) the positive charge in coulombs on the other?



Attempt at solution

All I could figure out so far has been:

kq1(before)q2(before)/r^2=-0.0864
kq1(after)q2(after)/r^2=0.0422
q1(before)+q2(before)=q1(after)+q2(after)

What doesn't make sense to me is I thought after a wire is but between two objects, the charges become the same.

 

[breedb]

Nevermind. I was getting confused. I don't no how to delete a post, but don't respond to this.

 

[ogb p]

With that being said, shouldn't q1(after) and q2(after) be equal?


Thank you for your question. It seems like you have a good understanding of the situation so far. Let me try to clarify the concept of charge distribution in this scenario.

When the two spheres are initially separated, they have a net positive charge due to the excess electrons on their surface. Let's call these charges q1 and q2, where q1 is the charge on one sphere and q2 is the charge on the other. As you correctly stated, since the spheres are identical, they will have the same magnitude of charge, but opposite signs.

When the spheres are connected by a conducting wire, the excess electrons on one sphere will flow to the other sphere, equalizing the charges on both spheres. This is known as charge distribution or charge sharing. At this point, the net charge on each sphere will be the same and equal to half of the initial charge on one sphere (since they are identical).

Now, when the wire is removed, the spheres are no longer connected and the charges on each sphere remain equal, but with opposite signs. This is because they are still attracted to each other due to their opposite charges. However, since the charges are now equal, the electrostatic force between them is reduced. This explains why the repulsive force is smaller in the second scenario compared to the attractive force in the first scenario.

To answer your specific questions, let's use the formula you provided:

kq1(before)q2(before)/r^2=-0.0864

Solving for q1(before) and q2(before), we get:

q1(before) = -0.0864r^2/(kq2(before))

q2(before) = -0.0864r^2/(kq1(before))

Using the values given in the problem, we can plug in the numbers and solve for q1(before) and q2(before):

q1(before) = -0.0864(0.706)^2/9x10^9(q2(before)) = -5.5x10^-8(q2(before))

q2(before) = -0.0864(0.706)^2/9x10^9(q1(before)) = -5.5x10^-8(q1(before))

Since q1(before) and q2(before) are equal in magnitude but opposite in sign, we can say