Force On Charges

[Air]

1. The problem statement, all variables and given/known data
A small object has charge Q. Charge q is removed from it and placed on a second small object. The two objects are placed 1 meter apart. For the force that each object exerts on each other to be a maximum, q should be:


2. Relevant equations
F=\frac{kQ_1Q_2}{r^2}


3. The attempt at a solution
The correct answer is \frac{Q}{2} although I can't seem to get to that. Is my formula selection correct. Also, if force is maximum, which value do I cosider it as. Thanks for the guidence in advance.

[gabbagabbahey]

1. The problem statement, all variables and given/known data
A small object has charge Q. Charge q is removed from it and placed on a second small object. The two objects are placed 1 meter apart. For the force that each object exerts on each other to be a maximum, q should be:


2. Relevant equations
F=\frac{kQ_1Q_2}{r^2}

Good; here r=1\text{m}, so you might as well write this as F=kQ_1Q_2 and not worry too much about the units.

Now, what are Q_1 and Q_2 for this problem (in terms of 'Q' and 'q')?


Also, if force is maximum, which value do I cosider it as. Thanks for the guidence in advance.

Well, you should have an equation for your force that is a function of 'q'; if I gave you some function of x and asked you to find the value of x that gives maximum value of that function, how would you do it? Apply the same idea to the force as a function of q.

[Air]

Well, you should have an equation for your force that is a function of 'q'; if I gave you some function of x and asked you to find the value of x that gives maximum value of that function, how would you do it? Apply the same idea to the force as a function of q.

It it was function of x, I would differentiate to find the maximum and equate it zero. How is that relevant?

EDIT: F = KQ^2\implies \frac{\mathrm{d}F}{\mathrm{d}Q} = 2KQ. Is this correct?

[gabbagabbahey]

It it was function of x, I would differentiate to find the maximum and equate it zero. How is that relevant?

Well, if you have a function of x you differentiate with respect to x and set it equal to zero.

So, if you have a function of q, why not differentiate it with respect to q and set it equal to zero? :wink:

[Air]

Well, if you have a function of x you differentiate with respect to x and set it equal to zero.

So, if you have a function of q, why not differentiate it with respect to q and set it equal to zero? :wink:

Ok, So if F=kQ_1Q_2, we can consider the charge to be F=kQ^2 hence as F = KQ^2\implies \frac{\mathrm{d}F}{\mathrm{d}Q} = 2KQ.

2KQ = 0 \implies Q=0. Hmmm... Is my error that I changed Q_1Q_2 to Q^2? :redface:

[gabbagabbahey]

EDIT: F = KQ^2\implies \frac{\mathrm{d}F}{\mathrm{d}Q} = 2KQ. Is this correct?

Right idea, but are Q_1 and Q_2 really both equal to Q?

If the first object starts out with charge Q and you take away an amount q how much is left? Call that Q_1.

If the second object starts out neutral and you add an amount of charge q, what is the charge on that object? Call that Q_2.

[Air]

Right idea, but are Q_1 and Q_2 really both equal to Q?

If the first object starts out with charge Q and you take away an amount q how much is left? Call that Q_1.

If the second object starts out neutral and you add an amount of charge q, what is the charge on that object? Call that Q_2.

F=k(Q-q)(q)

F=KQq - Kq^2

\frac{\mathrm{d}F}{\mathrm{d}q} = KQ - 2kq

When derivative equated to zero, max can be found: 0 = KQ - 2kq \implies KQ = 2Kq \therefore q = \frac{Q}{2}.

Woo, it works. I got the correct answer. Thanks a lot for the great help. You've been very patient and I appreciate the help. Thanks once again. :smile:

[gabbagabbahey]

F=k(Q-q)(q)

F=KQq - Kq^2

\frac{\mathrm{d}F}{\mathrm{d}q} = KQ - 2kq

When derivative equated to zero, max can be found: 0 = KQ - 2kq \implies KQ = 2Kq \therefore q = \frac{Q}{2}.

Woo, it works. I got the correct answer. Thanks a lot for the great help. You've been very patient and I appreciate the help. Thanks once again. :smile:

That's much better!:smile:

You should also check that the second derivative is negative so that you can be sure that you've found the maximum value and not the minimum value.

[Air]

That's much better!:smile:

You should also check that the second derivative is negative so that you can be sure that you've found the maximum value and not the minimum value.

Yes, \frac{\mathrm{d}^2F}{\mathrm{d}q^2} = -2q Which is negative hence it's a maximum.

[gabbagabbahey]

Looks good to me :approve: