# Applications of Gauss’s Law

## Applications of Gauss’s Law

According to Gauss’s law, the total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux through a closed surface is zero. If no charge is enclosed by the surface.

$${{\phi }_{E}}\,\,=\,\,\frac{{{Q}_{enclosed}}}{{{\varepsilon }_{0}}}$$.

Or

$${{\phi }_{E}}\,\,=\,\,\int\limits_{s}{\overrightarrow{E}}.\overrightarrow{da}$$.

So,

$${{\phi }_{E}}\,\,=\,\,\int\limits_{s}{\overrightarrow{E}}.\overrightarrow{da}\,\,=\,\,\frac{{{Q}_{enclosed}}}{{{\varepsilon }_{0}}}$$.

Here, $$E$$ is the total number of electric lines of force crossing through the given area, $$\phi$$ is the flux associated with the charge, $$Q$$ is the net charge enclosed by the closed surface, and $${{\varepsilon }_{0}}$$ is the permeability of free space. Gauss’s law describes how charges create electric fields.

Calculate the flux associated with a uniformly charged spherical shell.

$$\phi \,\,=\,\,\oint{E.ds}\,\,=\,\,E\left( 4\pi {{r}^{2}} \right)$$.

Applying Gauss’s law to find the electric field at a point located at a distance r outside the sphere.

$$E(4\pi {{r}^{2}})\,\,=\,\,\frac{Q}{{{\varepsilon }_{0}}}$$.

$$Electric\,\,Field(E)\,\,=\,\,\frac{Q}{4\pi {{r}^{2}}{{\varepsilon }_{0}}}$$.