Jatin Yadav's "Force of interaction"

Recently I solved an question of @Jatin Yadav Force of interection Click here

In this question I calculate Force of interaction by First Calculating Electric field due to long wire and then calculate force between them ( Which is Well explained in Jatin's Solution ). I had also solved that in this way First.

Doubt

In That Question If we consider Electric field due disc element in hemisphere and then Find The Force of interaction then How We can calculate This ?

My WORK

Consider an disc element on Hemisphere and small line element on wire as shown in figure . Sorry For poor Diagram But please Feel it Now I use standard result of electric field due to an uniformly charged disc that is

Eatx=σ2ϵ(1−xx2+R2)\displaystyle{{ E }_{ at\quad x }\quad =\quad \cfrac { \sigma }{ 2\epsilon } (1\quad -\quad \cfrac { x }{ \sqrt { { x }^{ 2 }+{ R }^{ 2 } } } )}.

F1,dueto2=∫dE2dq1(alongy−axis)∵dq1=λdy∵σ=ρRcos⁡θdθ∴dE2=ρRcos⁡θdθ2ϵ(1−y+Rsin⁡θ(y+Rsin⁡θ)2+(Rcos⁡θ)2)F1,dueto2=ρλR2ϵ∫y=0y=∞∫θ=0θ=π/2cos⁡θ(1−y+Rsin⁡θ(y+Rsin⁡θ)2+(Rcos⁡θ)2)dθdy\displaystyle{{ F }_{ 1,due\quad to\quad 2 }\quad =\quad \int { { dE }_{ 2 }{ dq }_{ 1 } } \quad \quad (\quad along\quad y-axis\quad )\\ \quad \quad \quad \\ \because \quad { dq }_{ 1 }\quad =\quad \lambda dy\\ \\ \because \quad \sigma =\quad \rho R\cos { \theta d\theta } \quad \quad \\ \\ \therefore \quad { dE }_{ 2 }\quad =\quad \cfrac { \quad \rho R\cos { \theta d\theta } }{ 2\epsilon } (1\quad -\quad \cfrac { y+R\sin { \theta } }{ \sqrt { { (y+R\sin { \theta } ) }^{ 2 }+{ (R\cos { \theta } ) }^{ 2 } } } )\\ \\ \boxed { { F }_{ 1,due\quad to\quad 2 }\quad =\quad \cfrac { \rho \lambda R }{ 2\epsilon } \int _{ y=0 }^{ y=\infty }{ \int _{ \theta =0 }^{ \theta ={ \pi }/{ 2 } }{ \cos { \theta (1\quad -\quad \cfrac { y+R\sin { \theta } }{ \sqrt { { (y+R\sin { \theta } ) }^{ 2 }+{ (R\cos { \theta } ) }^{ 2 } } } )d\theta } dy } } } }.

But From Here I get stuck

How can I find this integral ..??

Also When I use wolfram alpha then it says this this Integral not converges.!!

So please Tell Me where I'am wrong. ??

Thanks!!

#Calculus #ElectricityAndMagnetism #DoubleIntegrals #HelpMe!

Note by Deepanshu Gupta
6 years, 7 months ago

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1 vote

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Comments

This integral would give the same answer.

Note that ∫(1−y+Rsin⁡θ(y+Rsin⁡θ)2+R2cos⁡2θ)dy=y−(y+Rsin⁡θ)2+R2cos⁡2θ\displaystyle \int \bigg(1 - \dfrac{y+R \sin \theta}{(y+R \sin \theta)^2 + R^2 \cos^2 \theta}\bigg) {\mathrm dy} = y - \sqrt{(y+R \sin \theta)^2+R^2 \cos^2 \theta}

which approaches to Rsin⁡θR \sin \theta at y=∞y = \infty

⇒∫0∞(1−y+Rsin⁡θ(y+Rsin⁡θ)2+R2cos⁡2θdy)=R(1−sin⁡θ)\Rightarrow \displaystyle \int_{0}^{\infty} \bigg(1 - \dfrac{y+R \sin \theta}{\sqrt{(y+R \sin \theta)^2 + R^2 \cos^2 \theta}} {\mathrm dy}\bigg) = R(1- \sin \theta)

Hence, the integral can be expressed as:

ρλR2ϵ0∫0π/2R(1−sin⁡θ)cos⁡θdθ=ρλR24ϵ0\displaystyle \dfrac{\rho \lambda R}{2 \epsilon_{0} } \int_{0}^{\pi/2} R(1- \sin \theta) \cos \theta d \theta = \boxed{\dfrac{\rho \lambda R^2}{4 \epsilon_{0}}}

jatin yadav - 6 years, 7 months ago

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Wow !! Thanks a ton !! for Replying @jatin yadav

I haven't study about the double integration Yet But Now I understand Your Solution . So From your solution I draw The Conclusion That Double integral is nothing but it is calculate by integrating first integral keeping parameters of 2nd Integral Constant and then integrate 2nd Integral ,Is I'am Right ??

Also I would feel good if You will give me some facts about Double Integral (If Possible)

And Isn't There is mistake in typing of latex in starting line of your comment ?

Deepanshu Gupta - 6 years, 7 months ago

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Yes you are right , double integral is done in the same way as you said @Deepanshu Gupta

Lalit Pathak - 6 years, 6 months ago

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@Lalit Pathak – Thanks for replying! Lalit

Deepanshu Gupta - 6 years, 6 months ago
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