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Question-33450




Question Number 33450 by ajfour last updated on 16/Apr/18
Commented by ajfour last updated on 16/Apr/18
Find electric flux of a semiinfinite  thread of charge having linear  charge density λ, through a circular  disc of radius R if one end of the  thread is at the centre of this  circular area.
$${Find}\:{electric}\:{flux}\:{of}\:{a}\:{semiinfinite} \\ $$$${thread}\:{of}\:{charge}\:{having}\:{linear} \\ $$$${charge}\:{density}\:\lambda,\:{through}\:{a}\:{circular} \\ $$$${disc}\:{of}\:{radius}\:{R}\:{if}\:{one}\:{end}\:{of}\:{the} \\ $$$${thread}\:{is}\:{at}\:{the}\:{centre}\:{of}\:{this} \\ $$$${circular}\:{area}. \\ $$
Answered by 33 last updated on 18/Apr/18
consider an element  of lenght dx at dist x   from centre of  ring    dφ =  (dq/(2ε_0 ))(1−(x/( (√(R^2 +x^2 )))))  φ = ∫_(  0) ^(    L)   (((λdx))/(2ε_0 )) (1−(x/( (√(R^2  + x^2 )))))  ⇒ φ = (λ/(2ε_0  ))[x− (√(x^2  + R^2 )) ]_0 ^L   φ =  (λ/(2ε_0 ))lim_(L→∞) {L−(√(L^2 +R^2 )) +R}  φ = ((λR)/(2ε_0 ))
$${consider}\:{an}\:{element} \\ $$$${of}\:{lenght}\:{dx}\:{at}\:{dist}\:{x}\: \\ $$$${from}\:{centre}\:{of}\:\:{ring} \\ $$$$ \\ $$$${d}\phi\:=\:\:\frac{{dq}}{\mathrm{2}\epsilon_{\mathrm{0}} }\left(\mathrm{1}−\frac{{x}}{\:\sqrt{{R}^{\mathrm{2}} +{x}^{\mathrm{2}} }}\right) \\ $$$$\phi\:=\:\underset{\:\:\mathrm{0}} {\overset{\:\:\:\:{L}} {\int}}\:\:\frac{\left(\lambda{dx}\right)}{\mathrm{2}\epsilon_{\mathrm{0}} }\:\left(\mathrm{1}−\frac{{x}}{\:\sqrt{{R}^{\mathrm{2}} \:+\:{x}^{\mathrm{2}} }}\right) \\ $$$$\Rightarrow\:\phi\:=\:\frac{\lambda}{\mathrm{2}\epsilon_{\mathrm{0}} \:}\left[{x}−\:\sqrt{{x}^{\mathrm{2}} \:+\:{R}^{\mathrm{2}} }\:\right]_{\mathrm{0}} ^{{L}} \\ $$$$\phi\:=\:\:\frac{\lambda}{\mathrm{2}\epsilon_{\mathrm{0}} }{li}\underset{{L}\rightarrow\infty} {{m}}\left\{{L}−\sqrt{{L}^{\mathrm{2}} +{R}^{\mathrm{2}} }\:+{R}\right\} \\ $$$$\phi\:=\:\frac{\lambda{R}}{\mathrm{2}\epsilon_{\mathrm{0}} } \\ $$
Commented by ajfour last updated on 17/Apr/18
please explain the concept of  first line itself.  answer is correct.
$${please}\:{explain}\:{the}\:{concept}\:{of} \\ $$$${first}\:{line}\:{itself}. \\ $$$${answer}\:{is}\:{correct}. \\ $$
Commented by 33 last updated on 19/Apr/18
you are welcome sir :)
$$\left.{you}\:{are}\:{welcome}\:{sir}\::\right) \\ $$
Commented by 33 last updated on 18/Apr/18
for a point charge placed  at the axis of ring  at dist ′x′  flux through the disk can be  calculated     we can make an imaginary  curved surface (a part of sphere)with boundary  as the ring and its  centre coinciding with the   position of point charge.      φ = ( total flux) ((area of surface)/(area of sphere))
$${for}\:{a}\:{point}\:{charge}\:{placed} \\ $$$${at}\:{the}\:{axis}\:{of}\:{ring}\:\:{at}\:{dist}\:'{x}' \\ $$$${flux}\:{through}\:{the}\:{disk}\:{can}\:{be} \\ $$$${calculated} \\ $$$$\: \\ $$$${we}\:{can}\:{make}\:{an}\:{imaginary} \\ $$$${curved}\:{surface}\:\left({a}\:{part}\:{of}\:{sphere}\right){with}\:{boundary} \\ $$$${as}\:{the}\:{ring}\:{and}\:{its} \\ $$$${centre}\:{coinciding}\:{with}\:{the}\: \\ $$$${position}\:{of}\:{point}\:{charge}.\: \\ $$$$ \\ $$$$\:\phi\:=\:\left(\:{total}\:{flux}\right)\:\frac{{area}\:{of}\:{surface}}{{area}\:{of}\:{sphere}} \\ $$
Commented by 33 last updated on 18/Apr/18
Commented by 33 last updated on 18/Apr/18
x ≠ 0
$${x}\:\neq\:\mathrm{0} \\ $$
Commented by 33 last updated on 18/Apr/18
is that ok sir?
$${is}\:{that}\:{ok}\:{sir}? \\ $$
Commented by ajfour last updated on 18/Apr/18
thank you very much Sir.
$${thank}\:{you}\:{very}\:{much}\:{Sir}. \\ $$

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