Question

Find the flux through the surface $\left(APB\right)$ of the given Gaussian surface containing a point charge $+9\text{nC}$ as shown in figure.
[$ABC$ is an equilateral triangle and take ${ϵ}_o=9\times {10}^{-12}{\text{C}}^2/{\text{N-m}}^2$]

• A. $833{\text{N-m}}^2\text{/C}$
• B. $933{\text{N-m}}^2\text{/C}$
• C. $633{\text{N-m}}^2\text{/C}$
• D. $733{\text{N-m}}^2\text{/C}$

Answer 1

The correct option is B $933{\text{N-m}}^2\text{/C}$


The flat surface $\left(AB\right)$ of the given Gaussian surface can be assumed to be the base of a cone having semi-vertex angle $\theta$.
As triangle $ABC$ is an equilateral triangle, thus, $\angle ABC={60}^{\circ }$
and
$\theta =\frac{{60}^{\circ }}{2}={30}^{\circ }$.
The electric flux through the cone is given by
${\varphi }_{\text{cone}}=\frac{q}{2{ϵ}_0}(1-\cos \theta )$
The electric flux through the surface $APB$ is
${\varphi }_{\text{curved surface}}={\varphi }_{\text{container}}-{\varphi }_{\text{cone}}$
${\varphi }_{\text{curved surface}}=\frac{q}{{ϵ}_0}-\frac{q}{2{ϵ}_0}(1-\cos \theta )$
${\varphi }_{\text{curved surface}}=\frac{q}{2{ϵ}_0}+\frac{q\cos \theta }{2{ϵ}_0}$
${\varphi }_{\text{curved surface}}=\frac{q}{2{ϵ}_0}(1+\cos \theta )$ ... (1)
Putting the value of $\theta$ and $q$ in (1), we get
${\varphi }_{\text{curved surface}}=\frac{9\times {10}^{-9}}{2\times 9\times {10}^{-12}}(1+\cos {30}^{\circ })$
${\varphi }_{\text{curved surface}}=933{\text{N-m}}^2\text{/C}$
Hence, option (d) is correct.