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 \(\sigma\).

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

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

Details and Assumptions:

1) \( \sigma = 10^{-9} C , {\epsilon}_{0}=8.85 \times {10}^{-12} \text{ SI units} \)

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

\(\dfrac{r}{x} = \dfrac{\dfrac{a}{2}}{\dfrac{\sqrt{3}a}{2}}\)

\(\implies r= \dfrac{x}{\sqrt{3}}\) .

Now for electric field, We have,

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

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

We also know \(dq= \sigma r dx\).

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

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

Thanks for looking through my doubt.

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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 - 2 weeks, 5 days ago

@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 - 2 weeks, 5 days ago

Your way of writing fields due to an infinitismal element is wrong.Use a related result of finite length charged wire.\(E=\frac {k\lambda }{d}(\sin \theta _{1}+\sin \theta _{2})\) along the radial direction and\(\frac {k\lambda }{d}(\cos \theta _{1}-\cos \theta _{2})\) 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 - 2 weeks, 4 days ago

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

Md Zuhair - 1 week, 3 days ago

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 - 2 weeks, 5 days ago

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 – Did you do it by integrating the contributions from the strips?

Steven Chase - 1 week, 3 days ago

@Steven Chase – Yes sir. Like the way you told

@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.

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

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

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

@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 – Okay, Will try in that way! Thanks!

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Doubt on Ronak's Question.

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