Question

A spherical Gaussian surface of radius a = 400 cm surrounding a -50 µC charge. Find the
flux through

Answer 1

By definition, the flux of electric field is

$\Phi = E \cdot S$

Here E is a vector of electric field and S is a vector which module is equal to the area of the surface and direction is normal to the surface.

For the point-like charge and the sphere which surrounds it, the scalar product will $E \cdot S = |E| |S|$ since the vectors have the same direction.

$\displaystyle E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}$

$S = 4 \pi r^2$

$\displaystyle \Phi = ES = \frac{q}{\epsilon_0} = - \frac{50 \cdot 10^{-6}}{8.85 \cdot 10^{-12}}= 5.65 \cdot 10^6 \; \; [\frac{N \cdot m^2}{C}]$