Question

A point charge $q$ is placed on the vertex of a right circular cone. The semi vertical angle of the cone is ${60^ \circ }$ . Find flux of electric field through the base of the cone
A. $\dfrac{q}{{{\varepsilon _0}}}$
B. $\dfrac{q}{{2{\varepsilon _0}}}$
C. $\dfrac{q}{{3{\varepsilon _0}}}$
D. $\dfrac{q}{{4{\varepsilon _0}}}$

Answer 1

Hint- According to gauss law, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed , by the surface.
That is $\phi = \dfrac{q}{{{\varepsilon _0}}}$
Flux through the base of the cone is ${\phi _b} = \dfrac{A}{{{A_0}}}\dfrac{q}{{{\varepsilon _0}}}$
Where ${A_0}$ is the area of the whole sphere and $A$ is the area of the sphere below base of the cone
Area of a sphere ${A_0} = 4\pi {R^2}$
Area of the sphere below base of the cone \[A = 2\pi {R^2}\left( {1 - \cos \,\theta } \right)\]

Step by step solution:
According to gauss law, the net electric flux through any closed surface is equal to $\dfrac{1}{{{\varepsilon _0}}}$ times the total electric charge $q$ enclosed , by the
surface.
That is $\phi = \dfrac{q}{{{\varepsilon _0}}}$
Let us consider a sphere as a gaussian surface with its centre at the top of the cone and the slant height of the cone being the radius of the sphere.
Then flux through the whole sphere is $\phi = \dfrac{q}{{{\varepsilon _0}}}$ according to gauss law.
Flux through the base of the cone is ${\phi _b} = \dfrac{A}{{{A_0}}}\dfrac{q}{{{\varepsilon _0}}}$ ………………………...(1)
Where ${A_0}$ is the area of the whole sphere and $A$ is the area of the sphere below base of the cone
We know that area of a sphere is ${A_0} = 4\pi {R^2}$
\[
  dA = 2\pi r \times Rd\theta \\
   = 2\pi R\sin \theta \times Rd\theta \\
   = 2\pi {R^2}\sin \theta d\theta \\
 \]
since$r = R\sin \theta $
Integrate this from $0$to $\theta $
\[
  \int\limits_0^\theta {dA} = \int\limits_0^\theta {2\pi {R^2}\sin \theta d\theta } \\
  A = 2\pi {R^2}\left[ { - \cos \,\theta } \right]_0^\theta \\
  A = 2\pi {R^2}\left( {1 - \cos \,\theta } \right) \\
 \]
We have $\theta = {60^ \circ }$
Therefore,
\[
  A = 2\pi {R^2}\left( {1 - \cos \,\theta } \right) \\
   = 2\pi {R^2}\left( {1 - \cos {{60}^ \circ }} \right) \\
   = 2\pi {R^2}\left( {1 - \dfrac{1}{2}} \right) \\
   = \pi {R^2} \\
 \]
Substitute all the values in equation (1)
$
  {\phi _b} = \dfrac{{\pi {R^2}}}{{4\pi {R^2}}}\dfrac{q}{{{\varepsilon _0}}} \\
   = \dfrac{q}{{4{\varepsilon _0}}} \\
 $
This is the flux through the base of the cone.


So the answer is option D .

Note: This question can also be done by direct substitution of the values in the equation for flux through the base of a cone given by $\phi = \dfrac{q}{{2{\varepsilon _0}}}\left( {1 - \cos \,\theta } \right)$, where $\theta \,$is the semi vertical angle of the cone and $q$is the charge on the vertex