Applying coulomb's law

Question

Two identical objects have charges of 6*10^-6 C and -2.0 * 10^-6, respectively. When placed a distance d apart, their force of attraction is 2.0N. If the objects are touched together, then moved to a distance of separation of 2d, what will be the new force between them?

I got 0.5N, but thats wrong according to the answer sheet. Can someone show me how to do this? step by step?

thanks

Answer 1

The initial force is:

F = k q1 q2 / d^2

You can solve that to find the distance:

d = sqrt (k q1 q2 / F)

Then you touch them and they redistribute their charge so each has the same amount.

qf = (q1 + q2)/2

The new force will be:

Fnew = k qf qf / (2d)^2

= k (q1+q2)^2 / (16 d^2)

Plug in that original distance and get:

Fnew = k (q1+q2)^2 / 16(k q1 q2 / F)

= F (q1+q2)^2 / (16 q1 q2)

They give you the initial force (F) and charges (q1 and q2).  Plugnchug.

Note that since q1 and q2 have opposite sign, your answer comes out negative.  That's telling you that the new force is in the opposite direction of the original force.  It will be repulsive rather than attractive.

Answer 2

You may have forgotten that when the objects touch, their charges get added up. When they then get separated again, their charge becomes half of the total: 2*10^-6 C.

Answer 3

It's been forever since I've done this, so I might be a little rusty.

First, consider conservation of charge.  You have 6e-6 C and -2e-6 C, so add them and you get 4e-6 C total in the system.  When you touch them together, all of the electrons will rush to the positive one, resulting in equal charge on both of them.  Since it is equal on both, each has a charge of 2e-6 C.

Now, figure out the distance d. Kq1q2/d^2 yields that d is .23 m, I believe.  Redo this for the new charges and a d of .46 m.  I got .17 N.