# Min and max magnitude on pointlike charges

1. Jan 18, 2006

### Elysium

Hi, I'm having a problem with an exercise question on min and max electrical force.
I thought before that for the maximum magnitude, one of the points would have to be 'e' (fundamental unit of charge) and the other to be $$Q - e$$, since there would be an extreme amount of force exerted on the smaller force. For the minimum magnitude, thought both points would be $$\frac{Q}{2}$$. Does this make any sense?
I think that I went the wrong way in figuring this out since it's asking the max magnitude exterted on each other rather than a single point. I asumme that this question requires some calculus. Can anyone give me insight on how to figure this out?
Thank you.

2. Jan 18, 2006

### Staff: Mentor

For this problem I don't think that you need to worry about the fundamental charge, just assume a continuously varying charge. If you let the two charges be q1 and q2 then you have Q = q1 + q2 or q2 = Q - q1. Then write the electrostatic force in terms of q1 and q2, and do the substitution. You will have the force in terms of q1. Then find the q1 that maximizes the force.

Remember by Newton's 3rd law the force on q1 is the same magnitude as the force on q2.

-Dale

3. Jan 18, 2006

### Elysium

$$F_{elec} = \frac{1}{4\pi\varepsilon_0}\frac{|q_1||Q-q_1|}{r^2}$$
$$\frac{dF_{elec}}{dq_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q+2q_1}{r^2}$$
$$\frac{dF_{elec}}{dq_1} = 0 , q_1 = -Q$$
No, this isn't working... I didn't understand what you mean by the substitution. :(

4. Jan 18, 2006

### Elysium

So the magnitude of q1 and q2 remain equal regardless of the charge? Ok... Then with r constant, the maximum magnitude is when the charge approaches infinity? and for the minimum magnitude, the charge approaches 0.

Other than that, I have no idea.

5. Jan 18, 2006

### Staff: Mentor

Careful on your math. You did the substitution correctly, but you got a sign error. I don't know if the sign error came because of taking the absolute values. You are given that q1 and q2 are both positive, so you don't need to take the absolute values.

You should get Q/2 in the end.

-Dale

6. Jan 19, 2006

### Elysium

Correction:

$$F_{elec} = \frac{1}{4\pi\varepsilon_0}\frac{q_1(Q-q_1)}{r^2}$$
$$\frac{dF_{elec}}{dq_1} = \frac{1}{4\pi\varepsilon_0}\frac{Q-2q_1}{r^2}$$
$$\frac{dF_{elec}}{dq_1} = 0 , q_1 = \frac{Q}{2}$$

Ok, so this is the answer for the max, but what about the minimum? I'm still lost.

Last edited: Jan 19, 2006
7. Jan 19, 2006

### Gamma

Did you think this way for min charge. You place zero charge at one point and Q at the other. The minimum E. force is zero.

If there are two non zero charges placed a certain distance apart, the e. force between them is always non zero.

8. Jan 19, 2006

### Elysium

Dang, I got confused with another problem. I thought for a while both points had to have the same charge. :grumpy:

By E you mean the eletrical field?

Since they're positive charges, I gather that neither of them is zero. So that would leave a point with e charge and the other with Q-e charge for the minimum magnitude of force. Is that correct?

9. Jan 19, 2006

### Gamma

By E. Force I meant Electric Force. Not the electric field.

Yes, we found that both points should have the same charge (Q/2) to have maximum electric force. I think that's correct. I was answering your second question.

Zero is a number in math. :So why not e=0?

10. Jan 19, 2006

### Elysium

Because it's not in the set of positive numbers. A neutron (charge = 0) isn't a positive charge.

Thank you for your help (including DaleSpam). That problem has aggravated me for a little while. I did manage to finish to rest of the problems with little difficulty.

11. Jan 19, 2006

### Staff: Mentor

You are quite welcome!

For the min: 0 would be the minimum, but that is non-positive, so the minimum will be the smallest possible positive charge. So e is the second answer I think (elemental charge not exp).

I suspect that the question designer meant 0, but just worded the question poorly and said "positive" when they really meant "non-negative". That kind of thing happens all the time.

-Dale