electrostatic revision


Find the flux of the charge q through the upper cap of the cylinder....!!!.
(the charge q is kept on the bottom surface of the cylinder....)
assume the height to be h and the radius to be r

11 Answers

1
xYz ·

1
Unicorn--- Extinct!! ·

The answer is q2ε0*(1-h√h2+r2)

1
decoder ·

consider a strip of dx at a distance x.

the resultant field =2dEcosθ

now E=∫ 2dEcosθ

E= \int \frac{2KQ}{\sqrt{h^{2}+x^{2}}}\frac{h}{(h^{2}+x^{2})}

φ=E.dS

flux= \int \frac{2KQ}{\sqrt{h^{2}+x^{2}}}\frac{h}{(h^{2}+x^{2})} 2\pi xdx

now integrate and get the answer

1
Arshad ~Died~ ·

yes iota u are right....
@ XYZ u r wrong.....

1
Unicorn--- Extinct!! ·

My answer gives the flux through the cap.

23
qwerty ·

guyz pls see dis also
http://targetiit.com/iit-jee-forum/posts/polynomail-11969.html

sry for posting it here ..

1
Arshad ~Died~ ·

well flux through the cap is the answer dat i wanted....and that iota has written correctly.....

1
decoder ·

arshad iota's answer is correct then the same u will get by integrating my integral

\int_{0}^{R}{\frac{4h\pi kqxdx}{\sqrt[3]{h^{2}+x^{2}}}}

take h2+x2=t 2xdx=dt dx=dt/2x

\int_{0}^{R}{\frac{2h\Pi kqdt}{t^{3/2}}}

integrating u will get

\frac{q}{2\epsilon }(1-\frac{h}{\sqrt{h^{2}+r^{2}}})

1
Arshad ~Died~ ·

sorry decoder i didnt look closely at urs......yes dats right too....but wat xyz has done is wrong...

1
Unicorn--- Extinct!! ·

Decoder is correct, too. My method was different.

1
xYz ·

arshad evaluate my expression...
it also gives same.........btw i missed q in the expression

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