Question

a fixed circle is drawn of radius a and a charge Q is placed at a distance 3a/4 from the centre on a line through the centre perpendicular to the plane of the circle so that the flux of the electric field through the circle is 4π /5​

Answer 1

Given:

A fixed circle is drawn of radius 'a' and a charge Q is placed at a distance 3a/4 from the center on a line through the center perpendicular to the plane of the circle so that the flux of the electric field through the circle is 4π /5​ .

To Find:

Magnitude of charge Q is ?

Solution:

given;

flux through disc is ,   $\phi = \dfrac{4\pi}{5}$

Radius of circle is , R = a

Charge Q is placed at a distance 3a/4 , $d=\dfrac{3a}{4}$

Since, here according to question,

flux of the electric field through the circle given by ;

$\phi = \dfrac{Q}{2\epsilon_{0}}\left(1-\dfrac{d}{\sqrt{d^2+R^2} } \right)\\\\\\Q=\dfrac{\phi \times2\epsilon_{0}(\sqrt{d^2+R^2} )}{\sqrt{d^2+R^2}-d }$

therefore,  magnitude of point charge is given by,

$Q =\dfrac{ (4\pi/5)\times (2\times 8.85\times10^{-12}\times\sqrt{a^2} )}{\sqrt{a^2}-3a/4 }\\\\\\=\dfrac{32\pi \times 8.85 \times 10^{-12} \times a}{5a}\\\\= 56.64 \ \pi \times 10^{-12}C$