Sorry, But another Doubt! :P (Original)

Hey everyone! Sorry for disturbing again,I again have a doubt!

It will be very helpful, If you help me clearing it out.

Now lets jump into the question!


QUESTION

A cone is placed and a charge is placed at a distance of a m from the base of the cone as shown in the figure.

For our ease, Lets take a=1ma=1m , R=1mR=1m , h=1mh=1m ( height of the cone).

What we need to find is the flux due to the charge q through the cone!


MY SOLUTION

Here it goes, First of all, lets find the radius of the small disc we took,

Using Proportionality in triangles, we have

Rh=rhx\dfrac{R}{h} = \dfrac{r}{h-x}

    r=1x\implies r=1-x. This is the radius of the disc.

So the elemental area = 2π(1x)(dx) 2 \pi (1-x) (-dx).

Now Our flux is like this,

ϕ=SEcosθ×dS\phi = \displaystyle{\int_{S} E \cos \theta \times dS}

    ϕ=01q(1+x)×2π(1x)dx4πϵ0×(2(1+x2))3/2\implies \phi= \displaystyle{\int_{0}^{1} \dfrac{q(1+x) \times -2 \pi (1-x) dx}{4 \pi \epsilon_{0} \times (2(1+x^2))^{3/2}}}

    ϕ=01q(x21)dx42ϵ0(1+x2)3/2\implies \phi = \displaystyle{\int_{0}^{1} \dfrac{q ( x^2-1) dx}{4\sqrt{2} \epsilon_{0} (1+x^2)^{3/2}}}

After evaluating the integral, we get

    ϕ=q42ϵ02×arsinh(1)22\implies \phi = \dfrac{q}{4 \sqrt{2} \epsilon_{0}} \dfrac{\sqrt{2} \times arsinh(1)-2}{\sqrt{2}}

    ϕ=q42ϵ02×arsinh(1)22\boxed{\implies \phi = \dfrac{q}{4 \sqrt{2} \epsilon_{0}} \dfrac{\sqrt{2} \times arsinh(1)-2}{\sqrt{2}}}

But unfortunately our ϕ\phi comes out to be ve-ve.

Please help!!!

Thanks in Advance!

Note by Md Zuhair
1 year, 5 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Sort by:

Top Newest

@Steven Chase , @Aaron Jerry Ninan , @Thomas Jacob , @Shreyansh Mukhopadhyay , @Mark Hennings , @Ankit Kumar Jain

I guess you guys will try helping me finding the correct relation of ϕ\phi as I am getting -ve.

Thanks in Advance! I will be eagerly waiting for your response!

Md Zuhair - 1 year, 5 months ago

Log in to reply

Courtesy of Gauss' Theorem. the flux through the curved surface area of the cone is equal to the flux through its circular base CC. Thus Φ=  CEdS  =  q4πε0CrdSr3  =  q4πε00Ra(a2+u2)322πudu=  qa2ε0[(a2+u2)12]u=0R  =  qa2ε0(1a1a2+R2) \begin{aligned} \Phi & = \; \iint_C \mathbf{E} \cdot d\mathbf{S} \; = \; \frac{q}{4\pi \varepsilon_0} \iint_C \frac{\mathbf{r} \cdot d\mathbf{S}}{r^3} \; = \; \frac{q}{4\pi \varepsilon_0}\int_0^R \frac{a}{(a^2 + u^2)^{\frac32}}\,2\pi u\,du \\ & = \; \frac{qa}{2\varepsilon_0}\Big[-(a^2 + u^2)^{-\frac12}\Big]_{u=0}^R \; = \; \frac{qa}{2\varepsilon_0}\left(\frac{1}{a} - \frac{1}{\sqrt{a^2 + R^2}}\right) \end{aligned} and with the given parameters we have Φ  =  q2ε0(112) \Phi \; = \; \frac{q}{2\varepsilon_0}\left(1 - \tfrac{1}{\sqrt{2}}\right)

You are not calculating EdS\mathbf{E}\cdot d\mathbf{S} correctly. You are using EcosθdSE \,\cos\theta \,dS. In other words, you are assuming that dSd\mathbf{S} is horizontal (which works when integrating across the base CC), but this is not true when you are trying to integrate over the curved surface area of the cone.

Mark Hennings - 1 year, 5 months ago

Log in to reply

But if we are talkimg about the cylinders with same case of placement of charges then we see that they go by my method. Can you plz explain that.. as there also we can just find for base. Thanks.

Md Zuhair - 1 year, 5 months ago

Log in to reply

If we had a cylinder, the flux through the curved surface would be the flux through the near disc minus the flux through the far disc (so the total flux out of the cylinder is zero), and we can use my method again.

If you insist on integrating over the curved surface, the normal component of E\mathbf{E} would be EsinθE\sin\theta. You could integrate that over the surface.

Mark Hennings - 1 year, 5 months ago

Log in to reply

@Mark Hennings Oh I see.. So Esin theta dS will make it? But then what will be the dS? The same one?

Md Zuhair - 1 year, 5 months ago

Log in to reply

@Md Zuhair Is the cone solid or hollow?

Ankit Kumar Jain - 1 year, 5 months ago

Log in to reply

@Ankit Kumar Jain solid...

Md Zuhair - 1 year, 5 months ago

Log in to reply

@Md Zuhair Isn't the total flux through any closed surface due to some charge placed outside the body/surface =0 (Gauss Law)?

Ankit Kumar Jain - 1 year, 5 months ago

Log in to reply

@Ankit Kumar Jain yeah... but its the total. i want the curved SA flux

Md Zuhair - 1 year, 5 months ago

Log in to reply

@Md Zuhair dSdS would be the usual area element on a cylinder, yes. I should point out that one of the many benefits of Gauss' Theorem is that is allows us to avoid nasty integrals, and instead integrate over simpler surfaces. You are making life hard for yourself.

Mark Hennings - 1 year, 5 months ago

Log in to reply

@Mark Hennings True that sir. It is more clever by using Gauss Law rather than integrating. Thanks for your response! :D

Md Zuhair - 1 year, 5 months ago

Log in to reply

×

Problem Loading...

Note Loading...

Set Loading...