- #1
altegron
- 14
- 2
Homework Statement
http://img184.imageshack.us/img184/8125/chapter23number13qh8.png
Homework Equations
[tex] E = {k}_{e} \frac{q} {r^2}
[/tex]
The Attempt at a Solution
I figured that the electric field vector from the negative point charge would have to cancel out the electric field vector from the positive point charge so I said that 0 = E from the negative + E from the positive. I substituted values in and made the distance for the positive one r and the distance for the negative one (1-r). When I solved this for r, I get an answer like 1.03m, but the book says the answer is 1.82m to the left of the negative charge.
Even if I did the math wrong, what I am really wondering about it is the concept here. To me it seems like there should be 4 points where the field is 0. 1. At infinity, which the book mentions. 2. Somewhere far to the left of the negative charge, which I guess is the answer they put in the back. 3. Just slightly to the left of the negative charge. 4. And just slightly right of the negative charge.
My thinking on 3 and 4 is that E will get bigger and bigger as you get closer to the point (r->0, so E goes to infinity, right?). So, E will be large and positive close to the (+) charge, but as you go left to the (-) charge it will decrease to 0, and then become more negative as it reaches the (-) charge. Once it passes this, it will get bigger (from -infinity) until it equalizes with the (+) charge. Then, when you are rather far away it will be 0 again.
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