Electric Charge Problem with Unit Vectors.

1. Jul 19, 2007

PFStudent

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

Electric Charge Problem with Unit Vectors.

20. In Fig. 21-29, particles 1 and 2 of charge $q_{1} = q_{2} = +3.40{{.}}x{{.}}10^{-19}{{.}}C$ are on a y axis at distance d = 17.0 cm from the origin. Particle 3 of charge $q_{3} = +6.40{{.}}x{{.}}10^{-19}{{.}}C$ is moved gradually along the x axis from x = 0 to x = +5.0 cm. At what values of x will the magnitude of the electrostatic force on the third particle from the other two particles be
(a) minimum and
(b) maximum? What are the
(c) minimum and
(d) maximum magnitudes?

Figure 21-29,

2. Relevant equations

Coulomb’s Law,

Vector Form:
$$\vec{F}_{12} = \frac{k_{e}q_{1}q_{2}}{{r_{12}}^2}\hat{r}_{21}$$

Scalar Form:
$$|\vec{F}_{12}| = \frac{k_{e}|q_{1}||q_{2}|}{{r_{12}}^2}$$

3. The attempt at a solution

(a)

$$\Sigma \vec{F}_{3} = \vec{F}_{31} + \vec{F}_{32}$$

$$\Sigma \vec{F}_{3} = \frac{k_{e}q_{3}q_{1}}{{r_{31}}^{2}}\hat{r}_{13} + \frac{k_{e}q_{3}q_{2}}{{r_{32}}^{2}}\hat{r}_{23}$$

$$q_{1} = q_{2} = q$$

$$r_{31} = r_{32} = r_{3} = \sqrt{{x}^{2} + {d}^{2}}$$

$$\Sigma \vec{F}_{3} = \frac{k_{e}q_{3}\left(q\right)}{{\left(r_{3}\right)}^{2}}\hat{r}_{13} + \frac{k_{e}q_{3}\left(q\right)}{{ \left(r_{3}\right) }^{2}}\hat{r}_{23}$$

$$\Sigma \vec{F}_{3} = \frac{k_{e}q_{3}q}{{\left(\sqrt{{x}^{2}+{d}^{2}}\right) }^{2}}\left(\hat{r}_{13} + \hat{r}_{23}\right)$$

$$\Sigma \vec{F}_{3} = \frac{k_{e}q_{3}q}{{x}^{2}+{d}^{2}}\left(\hat{r}_{13} + \hat{r}_{23}\right)$$

$$\left|\Sigma \vec{F}_{3}\right| = \left|\frac{k_{e}q_{3}q}{{x}^{2}+{d}^{2}}\right|\left|\left(\hat{r}_{13} + \hat{r}_{23}\right)\right|$$

It is from here that I am stuck because I’m not sure exactly how to reduce the expression,

$$\left(\hat{r}_{13} + \hat{r}_{23}\right) = ?$$

I recognize that these are unit vectors and therefore have a magnitude of one (1), however I am still not sure as to how I should proceed.

Do I treat the addition of these unit vectors as normal vectors, noting that the magnitude is one?

I thought of doing that and described my attempt below.

Let,

$$\hat{r}_{13} + \hat{r}_{23} = \vec{r}_{13} + \vec{r}_{23}$$

And let,

$$\vec{r}_{13} + \vec{r}_{23} = \vec{r}_{3}$$

Then,

$$\vec{r}_{3} = \vec{r}_{3}_{x} + \vec{r}_{3}_{y}$$

Where it is noted that $\vec{r}_{3}_{y} = 0$, because: $\vec{r}_{13}_{y}$ and $\vec{r}_{23}_{y}$; cancel each other out.

So then,

$$\vec{r}_{3}_{x} = \vec{r}_{13}_{x} + \vec{r}_{23}_{x}$$

$${r}_{3}_{x}\hat{i} = {r}_{13}_{x}\hat{i} + {r}_{23}_{x}\hat{i}$$

$$\left|{r}_{3}_{x}\right| = \left|{r}_{13}_{x} + {r}_{23}_{x}\right|$$

$$\left|{r}_{3}_{x}\right| = \left|\left(+\frac{1}{\sqrt{2}}\right) + \left(+\frac{1}{\sqrt{2}}\right)\right|$$

$$\left|{r}_{3}_{x}\right| = \left|\sqrt{2}\right|$$

And so therefore,

$$\vec{r}_{3} = \vec{r}_{3}_{x} + \vec{r}_{3}_{y}$$

$$|\vec{r}_{3}| = \sqrt{{{r}_{3}_{x}}^{2} + {{r}_{3}_{y}}^{2}}$$

$$|\vec{r}_{3}| = \sqrt{{{r}_{3}_{x}}^{2} + (0)}$$

$$|\vec{r}_{3}| = \sqrt{{{r}_{3}_{x}}^{2}}$$

$$|\vec{r}_{3}| = |{r}_{3}_{x}|$$

$$|\vec{r}_{3}| = \sqrt{2}$$

Placing, this back in to the expression,

$$\left|\Sigma \vec{F}_{3}\right| = \left|\frac{k_{e}q_{3}q}{{x}^{2}+{d}^{2}}\right|\left|\left(\hat{r}_{13} + \hat{r}_{23}\right)\right|$$

Does not yield me the correct book answer….what am I doing wrong?

Any help is appreciated, thanks!

-PFStudent

2. Jul 20, 2007

chanvincent

Everything is correct up to this point:
and your problem is you do not know how to find the x component of the vector...
You must know the magnitude of the x component in the unit vector is not a constant, although the magnitude of the unit vector is.

the x component of $$\hat{r_{13}}$$ is
$$cos\theta = \frac{x}{\sqrt{x^2+d^2}}$$
the x component of $$\hat{r_{23}}$$ is the same,

so,
$$\left|{r}_{13}_{x} + {r}_{23}_{x}\right| = \frac{2x}{\sqrt{x^2+d^2}}$$

plug this in
$$\left|\Sigma \vec{F}_{3}\right| = \left|\frac{k_{e}q_{3}q}{{x}^{2}+{d}^{2}}\right| \left| \hat{r}_{13} + \hat{r}_{23} \right|$$

to get the $$\left| \Sigma \vec{F}_{3}\right|$$ in terms of x, then take its derivative with respect to x will yield its min/max value.... :tongue2: