Electric field and Plate (part 2)

Find the electric field in the z z direction due to a plate of area charge density λ \lambda at a distance a a above the plate.. B = 8 ° \angle B=8° If your answer comes in the form of E ( k ) = α λ ϵ 0 E(\vec{k})=\alpha \frac{\lambda}{\epsilon_{0}}
Find 100 α = ? 100 \alpha=?
Details and Assumptions
1) The triangle is isosceles .
2) A B = B C = 100 AB=BC=100
3) a = 0.5 a=0.5
4) ϵ 0 \epsilon_{0} is electric permittivity .
5) The triangle is in x y xy plane
6) Distance a a is measured from B B and is perpendicular to x y xy plane.
7) Cosider B as origin. Right side + x +x axis and in forward + y +y axis.


The answer is 1.111.

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1 solution

The electric field in z z due to a plane is (condition applied that a is small) E = λ 2 ϵ 0 E=\frac{\lambda}{2\epsilon_{0}}
For 360 ° = λ 2 ϵ 0 360°=\frac{\lambda}{2\epsilon _{0}}
Now for 8 ° = λ 90 ϵ 0 8°=\frac{\lambda}{90\epsilon_0}
E ( k ) = 10 9 E(\vec{k}) =\frac{10}{9}


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