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

$${please}\:{explain}\:{the}\:{concept}\:{of} \\ $$$${first}\:{line}\:{itself}. \\ $$$${answer}\:{is}\:{correct}. \\ $$

Commented by 33 last updated on 19/Apr/18

$$\left.{you}\:{are}\:{welcome}\:{sir}\::\right) \\ $$

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