electrostatic revision

The answer is q2ε₀*(1-h√h²+r²)

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

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

My answer gives the flux through the cap.

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

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}}})

Decoder is correct, too. My method was different.

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