# Homework Help: Optimization of Coulomb's Law

1. Sep 3, 2011

### eurekameh

Question:
Of the charge Q initially on a tiny sphere, a portion q is to be transferred to a second, nearby sphere. Both spheres can be treated a particles. For what value of q/Q>0.5 will the electrostatic force between the two parts have 1/5 of the maximum possible value?

Attempt:
F = [ k(Q-q)(q) ] / r^2
dF/dq = (k/r^2) d/dq(Qq - q^2)
= (k/r^2)(Q-2q) = 0
Q - 2q = 0
q = (1/2)Q --> Thus, this will give me the maximum electrostatic force between the two particles. However, the question is asking about the situation where it is 1/5 of the maximum value.

So,
F = [ k(Q-q)(q) ] / r^2, and using q = (1/2)Q or 2q = Q,
F/5 = [k(2q - q)(q)] / 5r^2
= (kq^2) / 5r^2 --> This is 1/5 of the maximum force.

Thus, to find q/Q:
F = [ k(Q-q)(q) ] / r^2 = (kq^2) / 5r^2 and I found q/Q to be 0.83333. This answer is wrong, however. Can anyone point me in the right direction?

2. Sep 3, 2011

### PeterO

Mixing symbols a bit

F = [ k(Q-q)(q) ] / r^2 should be 1/5 of k(Q/2)^2) / r^2

The force with the different charges, is only 1/5th of the force with the equal [half the original] charges

3. Sep 3, 2011

### Staff: Mentor

By writing 2q = Q you're forcing a particular value upon q rather than solving for a new one.

Suppose that fmax is the maximum force, and f is the force you're looking for. Then the desired condition is:
$$\frac{f}{f_{max}} = \frac{1}{5}$$
$$f = k \frac{(Q - q)(q)}{r^2}$$
and
$$f_{max} = k \frac{Q^2}{4 r^2}$$
Use both of these expressions to set up the ratio and solve for the 'new' q.

4. Sep 4, 2011

### eurekameh

I did F / Fmax and still got the answer to be 0.833333.

5. Sep 4, 2011