What Angle Should Bead 2 Be Positioned for Desired Electric Field?

[scoldham]

Homework Statement

The Figure below shows a plastic ring of radisu R = 50.0cm. Two small charged beads are on the ring. Bead 1 of charge +2.00 micro coulombs is fixed in the place at the left side of the ring. Bead 2 of charge +6.00 micro coulombs can be moved along the rignt. The two beads produce a net electric field of magnitude E at the center of the ring. At what (a) positive and (b) negative values of angle theta should bead 2 be positioned such that E = 2.00 x 10^5 N/C?

I do not have the picture in electronic format so I will describe it:

The figure is simply a circle on a xy coordinate system. Bead 1 is at the intersection of the x-axis and the circle on to the left of the origin. Bead 2 is on an arbitrary point on the circle in the second quadrant with a line of length R, from the origin, drawn to it. The angle, theta, is labeled as the angle between the line of length R and the x-axis.

Homework Equations

$$E = \frac{kQ}{R^2}$$

The Attempt at a Solution

The E-field induced by each charged bead is found using the above equation. I can calculate the affect bead 1 (the fixed bead) will have.

$$E_1 = \frac{kQ}{R^2}$$

Given the net E-field, I can determine the vector value for $$E_2$$ as $$E_1$$ and $$E_{net}$$ are in the same direction.

$$E_2 = E_{net} - E_1$$

Using $$E_2$$ form above:

$$E_2 = |E_2| (cos\vartheta + sin\vartheta)$$

At this point I get stuck... I'm not sure how to solve this for $$\vartheta$$

Did I do something wrong leading up to this... or is there some way to solve this that I'm not seeing?

 

[collinsmark]

Homework Equations

$$E = \frac{kQ}{R^2}$$

Directionality is very important in this problem. I would re-write the above equation to something of the form,

$$\vec E = \frac{kQ}{R^2} \hat r$$

where $\hat r$ is the unit direction vector from the charge to the test point.

The Attempt at a Solution

The E-field induced by each charged bead is found using the above equation. I can calculate the affect bead 1 (the fixed bead) will have.

$$E_1 = \frac{kQ}{R^2}$$

You need to be a little more specific here with your variables. There are two charges involved in the problem, so they should be labeled appropriately. Also, $\vec E_1$ has its own direction.

$$\vec E_1 = \frac{kQ_1}{R^2} \hat r_1$$

In this case it's pretty easy to express $\hat r_1$ in terms of $\hat x$ and $\hat y$. (Here, I am labeling Cartesian coordinate unit vectors as $\hat x$, $\hat y$ and $\hat z$. However, your textbook/coursework might use notation such as $$\hat i$$, $$\hat j$$ and $$\hat k$$, or maybe $\hat a_x$, $\hat a_y$ and $\hat a_z$. Whatever the case, you should find it easy to express $\hat r_1$ in terms of one of these Cartesian coordinate unit vectors.)

Given the net E-field, I can determine the vector value for $$E_2$$ as $$E_1$$ and $$E_{net}$$ are in the same direction.

Perhaps I am misunderstanding you here. $$\vec E_2$$, $$\vec E_1$$ and $$\vec E_{net}$$ are all in different directions.

$$E_2 = E_{net} - E_1$$

Using $$E_2$$ form above:

$$E_2 = |E_2| (cos\varthea + sin\vartheta)$$

I'm not sure what to make of that.

Form the equation for $$\vec E_2$$. Its unit vector will point in the direction of $\hat r_2$. The trick here is then to express [tex]\hat r_2 [/itex] in terms of $\hat x$ and $\hat y$. That's where the cosine and sines come in. Once you have the expression, you can (vector) sum it with $$\vec E_1$$ to get $$\vec E_{net}$$. <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f609.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":wink:" title="Wink :wink:" data-smilie="2"data-shortname=":wink:" /><br /> <br /> [Edit: Okay, now I think I see what you mean by $E_2 = |E_2| (cos \theta + sin\theta)$. But you need to include your Cartesian unit vectors in that equation (and there also might be '-' signs involved; remember the direction points from the charge Q₂ to the center of the ring).][/tex]

 

[scoldham]

Thanks for your help!

Wouldn't $$E_1$$ be in the $$-\hat i$$ direction?

Concerning $$E_2$$,

$$|E_2| (cos\vartheta -\hat i + sin\vartheta \hat j)$$

I would take these and add them (vector wise) for an equation with theta to solve for?

 

[collinsmark]

Thanks for your help!

Wouldn't $$E_1$$ be in the $$-\hat i$$ direction?

Oh, so close! http://www.websmileys.com/sm/sad/2.gif (that is, if I interpret your diagram description correctly, where Q₁ is to the left of the origin.)

I believe the direction is on the positive x-axis. Remember, the direction is from the charge to the test point (the test point is at the origin for this problem). So $$\hat r_1 = \hat i$$.

Concerning $$E_2$$,

$$|E_2| (cos\vartheta -\hat i + sin\vartheta \hat j)$$

I would take these and add them (vector wise) for an equation with theta to solve for?

The way I interpreted your description of the problem, is that the angle $\theta$ is with respect to the positive x-axis. In other words, if Q₂ is in the second quadrant, $\theta$ is between 90^o and 180^o. Is that right?

If so, try $$|E_2| (-cos\theta \hat i - sin\theta \hat j)$$

If I misinterpreted your description (such as the angle $\theta$ defined with respect to the negative x-axis, for example), just remember the direction is from the charge to the test location.

[Edit: Yes, once you have your vector sum, you should solve for $\theta$.]
[Another edit: Hint: The Pythagorean theorem may be involved in the process of solving for theta.]

 

[scoldham]

So I have:

$$E_{NET} = \frac{k}{R^2} [(Q_1 - Q_2 cos \vartheta) \hat i - Q_2 sin \vartheta \hat j]$$

I'm not sure how the pythagorean theorem comes into play though... $$sin^2 \vartheta + cos^2 \vartheta = 1$$

 

[collinsmark]

So I have:

$$E_{NET} = \frac{k}{R^2} [(Q_1 - Q_2 cos \vartheta) \hat i - Q_2 sin \vartheta \hat j]$$

Looks good to me!

[Edit: By the way, the above equation assumes that the angle $\vartheta$ starts from the positive x-axis. In other words, it means that if $\vartheta = 0$, then Q₂ is located at the point x = +0.5 m. This was my original assumption, based on your description of the figure. If my assumption is wrong, it will affect the above equation and the final answer to the problem!]

I'm not sure how the pythagorean theorem comes into play though... $$sin^2 \vartheta + cos^2 \vartheta = 1$$

The problem statement gave you the magnitude of $E_{net}$. So now you need to find the magnitude of the above expression. In other words, you need to find magnitude of a vector, where the vector is composed of two perpendicular components. (Hint: you need to find the "length" of the hypotenuse. ["length" is not really a good word here, since your solving for electric field strength, not distance. But I'm having difficulty thinking of a better word.])

 

[scoldham]

So... square the i and j components, sum them, and take the square root?

 

[collinsmark]

So... square the i and j components, sum them, and take the square root?

There you go! http://www.websmileys.com/sm/cool/049.gif

(By the way though, I just want to make sure I'm not misleading you. The equation in your previous post assumes that if $\vartheta = 0$ then Q₂ lies on the point x = +50.0 cm, y = 0. That's the picture I have in my head based on the original description. But I'm a little unsure of exactly how the angle $\vartheta$ is defined.)

 

[scoldham]

I got it at this point. Thanks for all your help.

And just for the record.. you were picturing it right.