Electrostatics: graphing/field vectors

[FritoTaco]

Homework Statement

(whole question/answer on IMG file also a scanned version in case)

Hello, this question is from my notes and I couldn't figure out the graphing part of it. We're supposed to use limits and see what happens to the equations as x goes to infinity and zero. My question (as you will find more below) concerns the right part of the graph that represents |E_{B}|

Just in case you can't see:
Two point charges are placed above. (in picture)
A = -7.00\times10^{-9} C
B = +2.00\times10^{-9} C
A is at (0,0)
B is at (3,0)

Questions

A.) Calculate E_{x} at (1,0).
B.) Sketch E_{x} vs. x for all x values.
C.) Clearly label any locations where E_{x}=0.
D.) Setup the calculation(s) to calculate the location(s) of any point(s) where E_{x}=0.
E.) Do the algebra to get the actual location(s).

Homework Equations

|E|=k\cdot\dfrac{|Q|}{r^2}

|E| is a vector.

k = 9.00\times10^{9}
Q = charge

The Attempt at a Solution

So, here I'm just going to show you how I'm evaluating the limits for |E_{A}| and |E_{B}|.

|E_{A}|=k\dfrac{Q_{A}}{x^2} as \lim\limits_{x \to 0} then k\dfrac{Q_{A}}{x}\rightarrow + \infty

As \lim\limits_{x \to \infty} then k\dfrac{Q_{A}}{x^2}\rightarrow 0


|E_{B}|=k\dfrac{Q_{A}}{(x-3)^2} as \lim\limits_{x \to 0} then k\dfrac{Q_{A}}{x}\rightarrow + \infty

As \lim\limits_{x \to \infty} then k\dfrac{Q_{A}}{(x-3)^2}\rightarrow 0


For the left part of the graph, this looks correct because as x increases, we get closer to zero, and as x decreases, we go to infinity. On the right side, we pass where E_{x} = 0 and where x > 3, we get a negative value. I don't see how to get there? If I attempt on a calculator, this how what I get:

|E_{B}|=\dfrac{9\times10^{9}(2\times10^{-9})}{100^2}=0.0018

So I used a random number greater than 3, which was 100. I get a small value which looks good, but it's not on the negative side of the axis like the graph shows. What am I missing?

 

[kuruman]

Why are you taking absolute values? I am not sure what you graphed, but I know what you should be graphing. That is
$$E_x=k \times 10^{-9}\left( \frac{-7}{x^2}+\frac{2}{(x-3)^2} \right)$$
Is that what you graphed?

 

[FritoTaco]

The absolute value bars were for finding magnitude. The direction is found separately unless you don't take absolute values. I graphed what I was taking the limits from (shown below), |E_{A}| is the left side of the graph and |E_{B}| is the right side.

|E_{A}| = \dfrac{k\cdot Q_{A}}{x^2}=\dfrac{9\times10^9(7\times10^{-9})}{100^2} (testing as limit as x goes to infinity by using 100) this gives me a small Ex value as x goes to infinity.
I used the same equation except now using as the limit goes to zero.
This gave me the left side of the graph.

And as I've shown at the end of my first post, I used limits for |E_{B}|

Ex=k×10−9(−7x2+2(x−3)2)

Did you get this equation without using the absolute value bar? You kept the negative sign for 7. I did use that equation but I didn't add them together, it was separated.

 

[kuruman]

You have to add them together as I showed you because $E_x$ is the x-component of the net electric field, i.e. the superposition of the electric fields generated by each charge.

 

[FritoTaco]

So I get E_{x}=k\times10^{-9}\dfrac{-5x^2+42x-63}{x^2(x-3)^2}

 

[kuruman]

So I get $E_{x}=k\times10^{-9}\dfrac{-5x^2+42x-63}{x^2(x-3)^2}.$

Sure, if you think that this expression is easier for you to see what's going on.

 

[FritoTaco]

Alright, but I don't see how I graph based on the equation?

 

[kuruman]

Alright, but I don't see how I graph based on the equation?

If you don't want to use a graphing calculator, spreadsheet or other such mathematical tool, you can do it ye olde fashioned way, the way I was taught in high school.
1. First find what your function does at very large +x and -x values.
2. Find the in-between points of interest. These are: (a) points where the function is zero; points where the function goes asymptotically to plus or minus infinity; points where the function has an extremum, maximum or minimum.
3. Connect the dots.

 

[FritoTaco]

Alright, thanks. I don't have time as of now so I will do it later!