# Electricity, Electic Force, and Charge Transfer

1. Feb 2, 2007

### Kaoi

Electricity, Electric Force, and Charge Transfer

1. The problem statement, all variables and given/known data

"Two conducting spheres have identical radii. Initially, they have charges of opposite sign and unequal magnitudes with the magnitude of the positive charge larger than the magnitude of the negative charge. They attract each other with a force of 0.224 N when separated by 0.4 m.

The spheres are suddenly connected with a thin wire, which is then removed. Now the spheres repel each other with a force of 0.039 N. What are the magnitudes of the initial positive and negative charges? Answer in units of C."

Given:
$$\bullet F_{e,i} = -0.224 N$$
$$\bullet r = 0.4 m$$
$$\bullet F_{e,f} = 0.039 N$$
$$\bullet k_{C} = 8.99 \times 10^9 \frac{N \cdot m^2}{C^2}$$

Unknown:
$$\bullet q_{pos} = ?$$
$$\bullet q_{neg} = ?$$

2. Relevant equations

$$F_{e} = \frac{k_{C}q_{1}q_{2}}{r^2}$$

3. The attempt at a solution

$$q_{pos} + q_{neg} = 2 q_{f}$$

$$F_{e,i} = \frac{k_{C}q_{pos}q_{neg}}{r^2}$$

$$F_{e,f} = \frac{k_{C}q_{f}^2}{r^2}$$

$$q_{f} = \sqrt{\frac{F_{e,f}r^2}{k_{C}}} = \frac {q_{pos} + q_{neg}}{2}$$

$$q_{pos}+q_{neg} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}$$

$$q_{pos} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}}- q_{neg}$$

At this point, I wasn't sure how to go about finding either of these charges, because you need one to solve for the other. Is there some clever mathematical trick or physical concept I'm missing here?

Last edited: Feb 2, 2007
2. Feb 2, 2007

### Kurdt

Staff Emeritus
From the original force equation (i.e. |F|=0.224) you can substitute for one value of the charge and obtain an equation with just a single unknown. Also this might need looking at:

$$q_{pos} + q_{neg} = 2 q_{f}$$

3. Feb 2, 2007

### Kaoi

If I solve that, I get:

$$F_{e,i} = \frac{k_{C}q_{pos}{q_{neg}}}{r^2}$$

$$q_{pos}q_{neg} = \frac{F_{e,i}r^2}{k_{C}}$$

And I can't solve for $$q_{pos}$$ and $$q_{neg}$$ at the same time...
I'm not sure I understand what you mean by "substitution" here...

Wouldn't the negative charge cancel out part of the positive charge?

$$q_{pos} + q_{neg} = q_{pos,remaining}$$

And then, since all that's left is positive charge, wouldn't the protons repel each other and spread evenly through the two connected spheres?

$$q_{f} = \frac{1}{2}q_{pos, remaining}$$

So:

$$2q_{f} = q_{pos,remaining} = q_{pos} + q_{neg}$$

At least, that's how I see it.

Last edited: Feb 2, 2007
4. Feb 2, 2007

### Kurdt

Staff Emeritus
Try

$$q_{pos} = \frac{F_{e,i}r^2}{k_{C}q_{neg}}$$

Then substitute into:

$$q_{pos} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}}- q_{neg}$$

and see how you go from there.

Last edited: Feb 2, 2007
5. Feb 2, 2007

### Kaoi

$$\frac{F_{e,i}r^2}{k_{C}q_{neg}} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}} - q_{neg}$$

$$q_{neg}+\frac{F_{e,i}r^2}{k_{C}q_{neg}} = 2r\sqrt{\frac{F_{e,f}}{k_{C}}}$$

$$q_{neg}^2 + \frac{F_{e,i}r^2}{k_{C}} = 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}}$$

$$q_{neg}^2 - 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} + \frac{F_{e,i}r^2}{k_{C}} = 0$$

So do I just solve that with the quadratic equation? (Sounds troublesome.)

6. Feb 2, 2007

### Kurdt

Staff Emeritus
I don't like this:

$$q_{pos} + q_{neg} = 2 q_{f}$$

You must remember sign conventions. If you fiddle with the signs you get a quadratic where you can complete the square.

7. Feb 2, 2007

### Kaoi

But if the second charge is negative, wouldn't subtracting it make it addition?

Last edited: Feb 2, 2007
8. Feb 2, 2007

### Kurdt

Staff Emeritus
All I was saying is you are adding a negative charge thus,

$$q_p+(-q_n) = q_p - q_n$$

Then your final equation will be solvable by the method of completing the square.

9. Feb 2, 2007

### Kaoi

Changing the signs, I get basically the same equation with a couple of signs reversed-- that still doesn't help me much...

$$q_{neg}^2 + 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} - \frac{F_{e,i}r^2}{k_{C}} = 0$$

That's a strange square to complete...

10. Feb 2, 2007

### Kurdt

Staff Emeritus
Have you ever heard of the completing the square method?

$$(x+a)^2 = x^2+2ax+a^2$$
$$x^2+2ax= (x+a)^2-a^2$$

Does the equation you have look familiar?

11. Feb 2, 2007

### Kaoi

So:

$$q_{neg}^2 + 2rq_{neg}\sqrt{\frac{F_{e,f}}{k_{C}}} - \frac{F_{e,i}r^2}{k_{C}} = 0$$

$$q_{neg}^2 + 2r\sqrt{\frac{F_{e,f}}{k_{C}}}q_{neg} = \frac{F_{e,i}r^2}{k_{C}}$$

$$x=q_{neg}$$, I know that, which would make:

$$q_{neg}^2 + 2q_{f}q_{neg} = (q_{neg} + q_{f})^2 - q_{f}^2$$

and that almost works, but $$F_{e,i}$$ and $$F_{e,f}$$ are two different quantities, so I can't use $$\sqrt{\frac{F_{e}r^2}{k_{C}}} = q_{f} = a$$, right? It looks like that's the only thing getting in the way. (This got a lot more complicated than I thought it would be...)

Last edited: Feb 2, 2007
12. Feb 2, 2007

### Kurdt

Staff Emeritus
You can use $$\sqrt{\frac{F_{e}r^2}{k_{C}}}$$ as a. Just apply the above relation that I gave. The other term in the equation just tags on the end unchanged.

13. Feb 2, 2007

### Kaoi

But the problem lies in which $$F_{e}$$ to use, since there are different $$F_{e}$$s on the sides of the equation, and they're not equal.

14. Feb 2, 2007

### Kurdt

Staff Emeritus
If we say $$\sqrt{\frac{F_{e}r^2}{k_{C}}}= k$$

Then,

$$q_{neg}^2 + 2q_{neg}k - \frac{F_{e,i}r^2}{k_{C}} = 0$$

$$(q_{neg}+k)^2 -k^2 - \frac{F_{e,i}r^2}{k_{C}} = 0$$

$$\left(q_{neg}+\sqrt{\frac{F_{e}r^2}{k_{C}}}\right)^2 - \frac{F_{e}r^2}{k_{C}} - \frac{F_{e,i}r^2}{k_{C}} = 0$$

15. Feb 2, 2007

### Kaoi

That clears it up a lot for me. Sorry for making you have to go to so much trouble.

So:

$$\left(q_{neg}+\sqrt{\frac{F_{e,f}r^2}{k_{C}}}\right) ^2 - \frac{F_{e,f}r^2}{k_{C}} - \frac{F_{e,i}r^2}{k_{C}} = 0$$

$$q_{neg} + \sqrt{\frac{F_{e,f}r^2}{k_{C}}} = \sqrt{\frac{F_{e,f}r^2}{k_{C}} + \frac{F_{e,i}r^2}{k_{C}}}$$

$$q_{neg} = \sqrt{\frac{F_{e,f}r^2 + F_{e,i}r^2}{k_{C}}} - \sqrt{\frac{F_{e,f}r^2}{k_{C}}}$$

And then it's just simple addition.

Physics is hard. :rofl:

Last edited: Feb 2, 2007
16. Feb 2, 2007

### Kurdt

Staff Emeritus
At least getting the other charge is not as hard. Physics ain't so bad just need to practise those basics such as quadratic equations and basic calculus etc. Looks like your algebra and manipulation is top notch though.