- #1
Bassalisk
- 947
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Homework Statement
Space distribution of electric charge is limited by to planes. Charge has the same intensity in planes parallel to these planes, but in dependence of the x coordinate, charge density is being distributed like:
[itex] \rho (x)=A\cdot x\cdot (d-x)[/itex] d is distance between those to planes, where A is a constant.
Find the electric field strength in all points in space.
Homework Equations
[itex]div \vec{E}=\frac{\rho }{\epsilon } [/itex]
And since we have only x coordinate, partial derivatives of y and z will be 0.
The Attempt at a Solution
[PLAIN]http://pokit.org/get/3da968a3cd6dab87fbe51fcfdcd6b116.jpg
Here are the solutions.
I get here:
[itex] |\vec{E}|= (2d-3x)\cdot \frac{A\cdot x^{2}}{6\cdot \epsilon}+C_{1} [/itex]
Then I used my intuition, which I think is WRONG. I found the stationary points of
[itex] \rho (x)=A\cdot x\cdot (d-x)[/itex] which resulted in d/2.
Then i figured if density is most dense or most thin there Electric field would be 0.
I got that constant you see in the picture [itex] C_{1}=- \frac{A\cdot d^{3}}{12\cdot \epsilon} [/itex]
What I don't get is how do we calculate the field outside the planes. Those solutions, are constants. But I can't figure out how did he get to that solution.
Any help?
P.S. Why are my LaTex fractions so small? How do I make them bigger?
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