- #1
FritoTaco
- 132
- 23
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 [itex]|E_{B}|[/itex]
Just in case you can't see:
Two point charges are placed above. (in picture)
[itex]A = -7.00\times10^{-9} C[/itex]
[itex]B = +2.00\times10^{-9} C[/itex]
A is at (0,0)
B is at (3,0)
Questions
A.) Calculate [itex]E_{x}[/itex] at [itex](1,0)[/itex].
B.) Sketch [itex]E_{x} vs. x[/itex] for all x values.
C.) Clearly label any locations where [itex]E_{x}=0[/itex].
D.) Setup the calculation(s) to calculate the location(s) of any point(s) where [itex]E_{x}=0[/itex].
E.) Do the algebra to get the actual location(s).
Homework Equations
[itex]|E|=k\cdot\dfrac{|Q|}{r^2}[/itex]
|E| is a vector.
[itex]k = 9.00\times10^{9}[/itex]
Q = charge
The Attempt at a Solution
So, here I'm just going to show you how I'm evaluating the limits for [itex]|E_{A}|[/itex] and [itex]|E_{B}|[/itex].
[itex]|E_{A}|=k\dfrac{Q_{A}}{x^2}[/itex] as [itex]\lim\limits_{x \to 0}[/itex] then [itex]k\dfrac{Q_{A}}{x}\rightarrow + \infty[/itex]
As [itex]\lim\limits_{x \to \infty}[/itex] then [itex]k\dfrac{Q_{A}}{x^2}\rightarrow 0[/itex]
[itex][/itex]
[itex]|E_{B}|=k\dfrac{Q_{A}}{(x-3)^2}[/itex] as [itex]\lim\limits_{x \to 0}[/itex] then [itex]k\dfrac{Q_{A}}{x}\rightarrow + \infty[/itex]
As [itex]\lim\limits_{x \to \infty}[/itex] then [itex]k\dfrac{Q_{A}}{(x-3)^2}\rightarrow 0[/itex]
[itex][/itex]
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 [itex]E_{x} = 0[/itex] 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:
[itex]|E_{B}|=\dfrac{9\times10^{9}(2\times10^{-9})}{100^2}=0.0018[/itex]
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?
Attachments
Last edited: