Question

Find the electric flux through the disc.

Answer 1

Hint: This question can be solved by concepts of Gauss Law. Connect all the points on the periphery of the disc to the point charge to generate a cone before proceeding to solve the problem.

Formula Used: The formulae used in the solution are given here.

$\oint E.dS={\displaystyle \frac{1}{{\epsilon }_0}}q$ where $q$ is the charge, $E$ is the electric field and $S$ represents the surface area.

Complete Step by Step Solution: 

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

Thus, the total flux linked with a closed surface is $1/{\epsilon }_0$ times the charge enclosed by the closed surface. $\oint E.ds={\displaystyle \frac{1}{{\epsilon }_0}}q$ where $q$ is the charge, $E$ is the electric field and $s$ represents the surface area. It has been given that a point charge $q$ is placed at a distance $d$ from the center of a circular disc of radius $r$.

If we connect the charge to all the points on the periphery of the disc, we get a cone.

This flux originates in a solid angle $4\pi$. In the given case the solid angle subtended by the cone subtended by the disc at the point charge is $\mathrm{\Omega }=2\pi \left(1-\cos \theta \right)$.So the flux of $q$ which is passing through the surface of the disc is, 

$\varphi ={\displaystyle \frac{q}{{\epsilon }_0}}{\displaystyle \frac{\mathrm{\Omega }}{4\pi }}={\displaystyle \frac{q}{2{\epsilon }_0}}\left(1-\cos \theta \right)$.

We get the value of $\theta$, from the figure.

$\cos \theta ={\displaystyle \frac{d}{\sqrt{d^2+r^2}}}$. Substituting the value of $\cos \theta$ in the equation for electric flux,

$\therefore \varphi ={\displaystyle \frac{q}{2{\epsilon }_0}}\left(1-{\displaystyle \frac{d}{\sqrt{d^2+r^2}}}\right)$. 

We get the electric flux as ${\displaystyle \frac{q}{2{\epsilon }_0}}\left(1-{\displaystyle \frac{d}{\sqrt{d^2+r^2}}}\right)$.

Note: The solid angle at the vertex of a cone can be derived by considering the spherical segment of the sphere centered at the vertex and passing through the periphery of the base and contained by the cone’s base, and integrating the small elemental solid angles. Gauss’s Law can be used to solve complex electrostatic problems involving unique symmetries like cylindrical, spherical or planar symmetry. Also, there are some cases in which calculation of electric fields is quite complex and involves tough integration. Gauss’s Law can be used to simplify the evaluation of electric fields in a simple way.