Question

. A spherical Gaussian surface of radius 1.00 m has a small hole of radius 10 cm. A point charge
of 2.00 × 10−9 C is placed at the center of this spherical surface. What is the flux through
this Gaussian surface?

Answer 1

Due to Gauss Law the electric field at any point of the sphere of radius  R ,  perpendicular to its surface and is equal in magnitude  equals:

$E=\frac{q}{4\times\pi\times e\times r^2};$

where e=const and $e=8.8\times 10^{-12};$

The flow through the spherical surface F is equal:

$F=E\times S;$

Where S is a sphere area:

$S=4\times \pi\times R^2;$

As a sphere has a hole, so we should minus its area from sphere area:

$F=E\times (S-s);$

$s=2\times \pi\times r^{2}\times (1-\cos \theta )$;

where r - radius of a hole;

So:

$F=\frac{q}{4\times \pi\times e\times r^2}\times (4\times \pi\times R^2-2\times \pi\times r^2\times (1-cos\theta)) = \frac{q}{4\times e\times r^2} \times (4\times R^2-2\times r^2);$

$F=\frac{2\times 10E-9}{4\times 8.8\times 10^{-12} \times 1^2} \times (4\times 1^2-2\times 0.1^2)=226.14=2.26*10^2;$