A Doubt on Electrostatics ( Inspired From Steven Sir)

You can check out this one first!.... https://brilliant.org/problems/focused-electric-field/?ref_id=1480760

QUESTION

Now my question is something different, Say we have a rectangular hyperbola, $x^2-y^2=a^2$ of which the right hand part is charged with linear charge density $-\sigma$ and the left one with $+ \sigma$ .

There is a Rod Lying on the y-axis, with mass $M$ , Lenght $2L$ and its center is kept at $(0,0)$ . It is charged with, say surface charge density, $+\lambda$ .

Check out the image.

This is it.

So We need to find the acceleration of the rod, as soon as we release it.

SOLUTION AS DONE BY ME (Not sure if correct).

If we draw the interaction between one element from the hyperbola to the rod, we can see that, The net force acting is in the X-direction and the Y- direction cuts off.

And The force from two opposite charged hyperbola on the rod will add up and will be like $dF_{x}=2 \times dF \sin \theta$ .

$\implies dF_{x}= \dfrac{2 \times dq_{1} \times dq_{2} \sin \theta }{(x^2+(y-l)^2)}$

Here (x,y) is the position of the charges on the hyperbola. and $l$ is the distance from the center of the rod to the element. and $\theta$ is the acute angle made by the rod with the line joining the element of rod and hyperbola.

$\implies dF_{x}= \dfrac{2 \times dq_{1} \times dq_{2} x }{(x^2+(y-l)^2)^{\tiny{\dfrac{3}{2}}}}$

We know $dq_{1} = \sigma dy \sqrt{1+(\dfrac{dx}{dy})^2}$ whereas dq = $\lambda dl$ ,

now putting these in the equation the force of interaction on the Rod comes out to be,

$\boxed{F=\displaystyle{\int_{-\infty}^{\infty} \int_{-L}^{+L} \dfrac{2 \lambda \sigma \sqrt{2y^2+a^2} dl dy}{(y^2+(y-l)^2+a^2)^{1.5}}}}$

$\boxed{A=\displaystyle{\int_{-\infty}^{\infty} \int_{-L}^{+L} \dfrac{2 \lambda \sigma \sqrt{2y^2+a^2} dl dy}{(y^2+(y-l)^2+a^2)^{1.5} \times M}}}$

I want you guys to verify if I am correct at every places and our expression comes same.

Thanks For Helping and Reading!

Question is original.

Please do the verification fast! I want to make a question on this and post it! And if you can do the integral, please tell me what comes.

Thanks again!

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

@Steven Chase , @Aaron Jerry Ninan , @Thomas Jacob , @Tapas Mazumdar , @Spandan Senapati , @David Mattingly , @Harsh Shrivastava , @Ranajay Medya.

Please help in this one! Thanks! I may have missed some names, you can add. Thanks

Md Zuhair - 3 years, 2 months ago

Hello. The approach looks right to me. I have two comments:

1) Under the radical in your numerator, doesn't it need to be $\large{\frac{2 y^2 + a^2}{y^2 + a^2}}$ ?
2) I'm not sure how to evaluate the integral formally, but it is no trouble to approximate numerically

Steven Chase - 3 years, 2 months ago

Never mind. When you bring in the x, it cancels the denominator. Looks good

Ya sir, I did that only, Okay, sir, Would you help me evaluate it by taking

$M=1 kg$ , $L=2m$ and $a=2m$ , $\lambda = 5 C/m$ and $\sigma= 10 C/m$

Then i will delete the doubt and post an question!

@Md Zuhair – Well, I spoke a little too soon when I said "no trouble". It's easy enough to write code to evaluate it. But then I would have to vary parameters (integration step size, integration window, etc.) and make sure that the convergence is good. I need to go to bed now, but I'd say you can give the numerical integration a shot and post it.

@Steven Chase – Okay sir! Will Try it! And which language do you use to do calculus in computer?

@Md Zuhair – Sounds good. I use Python. It's not as simple as calling an "integrate" function, but you can easily do a Riemann sum (or a double Riemann sum).

@Steven Chase – Oh nice, I see... Sir according to https://www.integral-calculator.com/ , I got 410.372632 . Should I try Posting it?

@Md Zuhair – If you are confident in the answer, go ahead and post it. I'll try it tomorrow. It's a neat problem

@Steven Chase – Okay sir! Sure, Lets post, if reports come, I will delete it and correct it. Thanks for Helping me! I guess in US, it night! So goodnight sir! Here , In India, Its 1:33 pm :)

@Md Zuhair – @Md Zuhair Hiya!! I see that you are so much advanced in physics.....That's OSM..!! Do you mind if I ask that where do u learn it?? I am referring to Resnick Halliday, and have only finished the first 7-8 chapters......(Mechanics excluding rotation).....But I need to improve my skills.....any tips??

Aaghaz Mahajan - 3 years, 2 months ago

@Aaghaz Mahajan – Arre! Waise nai hai. Ok, So I do learn from Books, Youtube channels and Brilliant (90%) and learn many new concepts on solving problems. It clears out my concepts, Actually, Do not ask, Where did you learn, Ask From where not to learn, You should try learning from everywhere. Kuch zyada hi bol diya, Lagta Hai. Ok.. So Resnick Halliday is good, and try learning Rotation fast, it still is difficult to me in some respects!

Thanks :D!

@Md Zuhair – Ohhh I see.....zyaada nahi bola hai.....!! I am more into Maths than any other subject aur ab mujhe Phy and Chem bhi improve karni hain....( I have not even started the NCERT of Chem....!! well except that atomic structure waala chapter!! )....Ok....then I'll first finish off Resnick!!
Thanks to u man!!

@Aaghaz Mahajan – Arre .. no thanks. Best of luck

@Md Zuhair – I posted some example code under the one with the tension and the bob.

https://brilliant.org/problems/breaking-string/

@Steven Chase – Okay! Will check it out

@Chew-Seong Cheong , Sir, Can you make this look prettier?

https://brilliant.org/problems/complex-harmonic-motion/?ref_id=1484330

Bro !! Abhi bhi online!! Sorry I won't try ur problem....

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

A Doubt on Electrostatics ( Inspired From Steven Sir)

A=∫−∞∞∫−L+L2λσ2y2+a2dldy(y2+(y−l)2+a2)1.5×M\boxed{A=\displaystyle{\int_{-\infty}^{\infty} \int_{-L}^{+L} \dfrac{2 \lambda \sigma \sqrt{2y^2+a^2} dl dy}{(y^2+(y-l)^2+a^2)^{1.5} \times M}}}A=∫−∞∞​∫−L+L​(y2+(y−l)2+a2)1.5×M2λσ2y2+a2​dldy​​

Comments

$\boxed{A=\displaystyle{\int_{-\infty}^{\infty} \int_{-L}^{+L} \dfrac{2 \lambda \sigma \sqrt{2y^2+a^2} dl dy}{(y^2+(y-l)^2+a^2)^{1.5} \times M}}}$