Doubt on Ronak's Question.

https://brilliant.org/problems/electric-fields-due-to-different-kinds-of-plates/.

The link is given.

Here, while doing with Integration results in different value. I will be glad if someone points out where i am wrong.


QUESTION

We have a triangular uniformly charged plate of charge density σ\sigma.

We take a point just above a vertex of the triangular plate at a distance dd perpendicular to the plane of the triangle. Let the vertical component of electric field at that point be E(d)E(d) in V/mV/m

Find limd0E(d) \displaystyle \lim_{d \rightarrow 0}{E(d)}

Details and Assumptions:

1) σ=109C,ϵ0=8.85×1012 SI units \sigma = 10^{-9} C , {\epsilon}_{0}=8.85 \times {10}^{-12} \text{ SI units}

2) The triangle is equilateral. By vertical component of electric field I mean electric field perpendicular to the plane of charge.

3) The plate is non conducting.


MY SOLUTION

So By using Proportionality in triangle,

We get

rx=a23a2\dfrac{r}{x} = \dfrac{\dfrac{a}{2}}{\dfrac{\sqrt{3}a}{2}}

    r=x3\implies r= \dfrac{x}{\sqrt{3}} .

Now for electric field, We have,

dEsinθ=kd(dq)(x+d)32dE \sin\theta = \dfrac{kd (dq)}{(x+d)^{\tiny \dfrac{3}{2}}}

Ey=03a/2kd(dq)(x+d)32E_{y} = \displaystyle{\int_{0}^{\sqrt{3}a/2} \dfrac{kd (dq)}{(x+d)^{\tiny \dfrac{3}{2}}}}

We also know dq=σrdxdq= \sigma r dx.

After solving, we get limd0E(d)=10.39\displaystyle{\lim_{d \rightarrow 0} E(d)= 10.39}.

If I am wrong anywhere, Please help me correct it. Also the integral calculation can be a error.

Thanks for looking through my doubt.

Note by Md Zuhair
1 year, 4 months ago

No vote yet
1 vote

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Comments

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@Steven Chase @Mark Hennings @Thomas Jacob @Aaron Jerry Ninan @Ranajay Medya @Spandan Senapati .

Please help me! I may have skipped some names. You can add them too!

Md Zuhair - 1 year, 4 months ago

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I just solved it by integrating, with no simplifying tricks like the one given in the solution (ingenious as it is). I'll put up a solution in the next few days.

Steven Chase - 1 year, 4 months ago

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That means my integral is correct, But the value isnt?

Md Zuhair - 1 year, 4 months ago

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@Md Zuhair I used an integral, but it wasn't necessarily your integral.

Steven Chase - 1 year, 4 months ago

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@Steven Chase Ohkay, I see. I dont know why, I am not getting it. Did you just saw my solution. You can point out any mistake if you get there. else it may be fine.

Md Zuhair - 1 year, 4 months ago

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@Md Zuhair When calculating the contribution from one of the strips, you have to integrate as well. So the problem effectively boils down to a double integral. But I did the two integrations sequentially.

Steven Chase - 1 year, 4 months ago

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@Steven Chase Okay, So I have to take a small element in that strip too?

Md Zuhair - 1 year, 4 months ago

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@Md Zuhair Yes. So integrating with infinitesimals within the strip gives you an expression for the contribution of a strip. Then integrating the contributions from the strips gives the final result. The first integral is readily done with an integral table. For the second integral, I used Wolfram Alpha. Then you have some interesting philosophizing to do regarding the limit of an arctangent before getting to the final answer.

Steven Chase - 1 year, 4 months ago

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@Steven Chase Okay, Will try in that way! Thanks!

Md Zuhair - 1 year, 4 months ago

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Thanks sir! Finally solved it using Integrals! But, Why do i need double integral here? Why cant i apply integral from the middle of the thin strip in the triangle?

Md Zuhair - 1 year, 4 months ago

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@Md Zuhair Did you do it by integrating the contributions from the strips?

Steven Chase - 1 year, 4 months ago

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@Steven Chase Yes sir. Like the way you told

Md Zuhair - 1 year, 4 months ago

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@Md Zuhair Ok. And then it also takes an integral to figure out the contribution from a strip. That's why I refer to the whole process as effectively being a double integral.

Steven Chase - 1 year, 4 months ago

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Your way of writing fields due to an infinitismal element is wrong.Use a related result of finite length charged wire.E=kλd(sinθ1+sinθ2)E=\frac {k\lambda }{d}(\sin \theta _{1}+\sin \theta _{2}) along the radial direction andkλd(cosθ1cosθ2)\frac {k\lambda }{d}(\cos \theta _{1}-\cos \theta _{2}) along the perpendicular. But if you want to develop your skills avoid integrals,symmetry helps a no of times,like considering hexagonal plates.

Spandan Senapati - 1 year, 4 months ago

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I have a small qs. If does that σ2ϵ0\dfrac{\sigma}{2 \epsilon_{0}} can be used for Any kind of plates? As i knew it is used for rectangular plates. Thanks

Md Zuhair - 1 year, 4 months ago

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In this Q the concept of relative largeness plays a very important role. d->0 indirectly states that the point is near an infinitely large plane. I'll check if we can arrive at the same conclusion using the integration process.

Ranajay Medya - 1 year, 4 months ago

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