Question

The given figure shows a spherical Gaussian surface and a charge distribution. When calculating the flux of electric field through the Gaussian surface, the electric field will be due to

• A. $+q_3$ alone
• B. $+q_1$ and $+q_3$
• C. $+q_1,+q_3$ and $-q_2$
• D. $+q_1$ and $-q_2$

Answer 1

Answer: (D)

Shown a spherical Gaussian surface and a charge distribution.
The flux through the opherical Gaussian surface will be due to charges $+q_1$ and $-q_2$ only. Because, $q_3$ is outride surface.
So flux $=\varphi =\frac{\text{ net charge enclosed }}{{\epsilon }_0}$
$\int Eds=\frac{\left(q_1-q_2\right)}{{\epsilon }_0}$
$E\left[4\pi r^2\right)=\frac{q_1-q_2}{{\epsilon }_0}$
[$r=$ radius of sphere]
$E=\frac{\left(q_1-q_2\right)}{{\epsilon }_0}\left[\frac{1}{r^2}\right]$
We can see that $E\alpha \left(q_1-q_2\right)$.
Thus, the electric field, while calculating flux through gaussian surface will be due to charges $+q_1$ and $-q_2$ only.