Two Conducting Spheres Problem - Applying

[PFStudent]

Homework Statement

9. Two identical conducting spheres, fixed in place attract each other with an electrostatic forces of 0.108 N when their center-to-center separation is 50.0 cm. The spheres are then connected by a thin conducting wire. When the wire is removed, the spheres repel each other with an electrostatic forces of 0.0360 N. Of the initial charges on the spheres, with a positive net charge, what was
(a) the negative charge on one of the, and
(b) the positive charge on the other?

Homework Equations

$$\vec{F}_{12} = \frac{kq_{1}q_{2}}{r^2}\hat{r_{21}}$$

The Attempt at a Solution

I know I can use Coulomb's law to solve this problem, however I want to just make sure I first understand the vector form of it.

So, the law

$$\vec{F}_{12} = \frac{kq_{1}q_{2}}{r^2}\hat{r_{21}}$$

basically says that q1 and q2 are treated as positive where the: negative sign for attraction or positive sign for repulsion; comes from the radial hat vector right?

Ok, so for example in this problem, we have two charges q1 and q2:

http://img162.imageshack.us/img162/1843/91rg6.png

Ok, so using the vector form of coulomb's law I treat q1 and q2 as positive charges and the negative sign for attraction should come from the radial vector...but how?

And since this is a radial vector the direction of $\hat{r_{21}}$ is that: the distance beginning at q2 to q1 or beginning at q1 to q2?

 

[Doc Al]

So, the law

$$\vec{F}_{12} = \frac{kq_{1}q_{2}}{r^2}\hat{r_{21}}$$

basically says that q1 and q2 are treated as positive where the: negative sign for attraction or positive sign for repulsion; comes from the radial hat vector right?

No. The sign comes from the product of the two charges, which can be positive or negative. If the charges are opposite, you'll have a negative sign--which means the direction is opposite to radial unit vector and thus towards the other charge (in other words, attractive).

Ok, so using the vector form of coulomb's law I treat q1 and q2 as positive charges and the negative sign for attraction should come from the radial vector...but how?

Treat the charges as they are: one positive, one negative.

And since this is a radial vector the direction of $\hat{r_{21}}$ is that: the distance beginning at q2 to q1 or beginning at q1 to q2?

It stands for a unit vector pointing from charge 2 towards charge 1. Thus if the charges had the same sign, the direction of the force would be repulsive.

 

[PFStudent]

Hey,

Doc Al thanks for the information.

Ok so proceeding to the solution.

Coulomb's Law:

$$\vec{F}_{12} = \frac{kq_{1}q_{2}}{r^2}\hat{r}_{12}$$

Can be rewritten with the scalar component $F_{12}$

$$\vec{F}_{12} = F_{12}\hat{r}_{21}$$

Also it is known that q1 and q2 are unlike-sign to account for the initial attraction.

Therefore, letting $F_{12} = 0.108 N \equiv +$.

and $F'_{12} = 0.0360 N \equiv +$. We have:

$$-{F}_{12} = \frac{k_{e}q_{1}q_{2}}{r^2}$$
$${F'}_{12} = \frac{k_{e}{q'}_{1}{q'}_{2}}{r^2}$$

Where ${q'}_{1}$ and ${q'}_{2}$ represent the charges after the wire is removed. And so due to Conservation of Charge.

$${q'}_{1} = {q'}_{2} = \left(\frac{{q}_{1} + {q}_{2}}{2}\right)$$

And therefore,

$${F'}_{12} = \frac{k_{e}(q_{1}+q_{2})^2}{4{r^2}}$$

In solving the above for $(q_{1}+q_{2})$, the problem that arises is whether to choose the positive or negative root to solve for q1 and q2. And justifying that choice.

$$q_{1}+q_{2} = \pm\sqrt{\frac{4{r^2}{{F'}_{12}}}{{k}_{e}}}$$

(Note: Either root will lead to the same values, however signs will be reversed and because part (a) explicitly asks for the negative charge, then only either one of the two possible roots may be used to solve for the unknowns, q1 and q2.)

Any ideas on this?

Thanks!

-PFStudent

 

[Doc Al]

Recall that the net charge on both spheres is positive. What does that tell you about $(q_{1}+q_{2})$?

 

[PFStudent]

Hey,

Thanks Doc Al for the help, and because the original problem specifies that the net charge on both spheres is positive, this mathematically means that the sum of the two charges is greater than or equal to zero.

Or in other words,

$$(q_{1}+q_{2}) \geq 0$$

Which implies that only the positive root of,

$$q_{1}+q_{2} = \pm\sqrt{\frac{4{r^2}{{F'}_{12}}}{{k}_{e}}}$$

$$\therefore$$

$$q_{1}+q_{2} = +\sqrt{\frac{4{r^2}{{F'}_{12}}}{{k}_{e}}}$$

will lead to the correct values for the (a) negative charge and (b) positive charge.

Man, I learned a lot from this problem, :)

Thanks for the help Doc Al!

-PFStudent

P.S. Sorry for the late response.

 

[Doc Al]

Excellent. I am delighted that you stuck with it. Good work.