Electric Charge Problem with Unit Vectors.

[PFStudent]

Homework Statement

Electric Charge Problem with Unit Vectors.

20. In Fig. 21-29, particles 1 and 2 of charge $q_{1} = q_{2} = +3.40{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}10^{-19}{\textcolor[rgb]{1.00,1.00,1.00}{.}}C$ are on a y-axis at distance d = 17.0 cm from the origin. Particle 3 of charge $q_{3} = +6.40{\textcolor[rgb]{1.00,1.00,1.00}{.}}x{\textcolor[rgb]{1.00,1.00,1.00}{.}}10^{-19}{\textcolor[rgb]{1.00,1.00,1.00}{.}}C$ 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,

http://img530.imageshack.us/img530/2170/phy111xu0.png

Homework Equations

Coulomb’s Law,

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

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

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 into 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

 

[chanvincent]

Everything is correct up to this point:

$$\left|{r}_{3}_{x}\right| = \left|{r}_{13}_{x} + {r}_{23}_{x}\right|$$
$$\left|{r}_{3}_{x}\right| = \left|\left(+\frac{1}{\sqrt{2}}\right) + \left(+\frac{1}{\sqrt{2}}\right)\right|$$

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...

 

[m2003]

As a scientist, it's important to approach problems in a systematic and logical manner. Your attempt at solving the problem shows that you understand the basics of Coulomb's law and vector addition, but there are a few errors in your calculations.

First, when adding unit vectors, you can't simply add their magnitudes. You need to add their components in each direction (x and y in this case) and then find the magnitude of the resultant vector. So, the correct expression for the sum of unit vectors is:

|∑F3| = |kq3q|/(x2+d2) * √(r13x + r23x)^2 + (r13y + r23y)^2)

Second, your calculation for r3x is incorrect. The x component of r3 should be the difference between the x components of r13 and r23, not their sum. So, it should be:

r3x = r13x - r23x

Also, the x component of r13 and r23 should be calculated using trigonometry, since they are at an angle of 45 degrees from the x-axis. So, r13x and r23x should be:

r13x = r13 * cos(45 degrees)

r23x = r23 * cos(45 degrees)

Finally, when you plug in the values for r13 and r23, you should get the correct answer for the minimum and maximum values of the electrostatic force on particle 3.

In summary, the key to solving this problem is to carefully calculate the components of r13 and r23, and then use vector addition and the magnitude formula to find the resultant force. Remember to pay attention to the direction of the vectors and use trigonometry when necessary. Keep practicing and you'll become more comfortable with these types of problems. Good luck!