Question

a point charge q is placed at a perpendicular distance x from centre of very large circular sheet .find flux through sheet .R is radius of sheet

Answer 1

Answer:

180 -90 is eqaual to... the answer is 90

Answer 2

Answer:

Electric flux through the circular sheet of radius R is ${\frac{q}{2\epsilon _{0} } } (1-\frac{x}{\sqrt[]{x^{2} +R^{2} }})$

Explanation:

Electric flux is given by $\phi=\int\ {\bar E} \, d\bar s$

                                         $={ E} \, dscos\theta$

Consider a small element in the shape of ring of radius $r$ and hence width is $dr$.

Then electric field is in radial direction, given by $E= {\frac{Kq}{x^{2} +r^{2} }$,

areal vector is perpendicular to the sheet, given by $ds=2\pi rdr$

and the angle is $cos\theta=\frac{x}{\sqrt[]{x^{2} +r^{2} }} }$

Substituting between the limits 0 and radius of sheet R, the total flux through the element is

               $\phi_{total} =\int\ {E} \, ds cos\theta$

                        $=\int\limits^0_R {\frac{Kq}{x^{2} +r^{2} } 2\pi rdr\frac{x}{\sqrt[]{x^{2} +r^{2} }} }$

                        $=Kq\pi x \int\limits^R_0 { \frac{2 rdr}{\sqrt[]{x^{2} +r^{2} }} }$

Let ${x^{2} +r^{2} }} }=u$ and $2rdr=du$

Then when $r=0\implies x^{2} =u$ and $r=R\implies {x^{2} +R^{2} }} }=u$

Substituting the limits,

                               $\phi_{total} = {\frac{q}{2\epsilon _{0} } } (1-\frac{x}{\sqrt[]{x^{2} +R^{2} }})$