Question
State and Prove Gauss theorem in electrostatics.
✓
Answer
Statement: The net-outward normal electric flux through any closed surface of any shape is equal to $\frac{1}{\epsilon_0}$ times the total charge contained within that surface, $\frac{1}{\epsilon_0}$ i.e., $\oint\text{S }\vec{\text{E}}.\vec{\text{ds}}=\frac{1}{\epsilon_0}\Sigma\text{q}$ Where $\oint\limits_{\text{S}}$ indicates the surface integral over the whole of the closed surface, $\Sigma\text{q}$ Is the algebraic sum of all the charges (i.e., net charge in coulombs) enclosed by surface S and remain unchanged with the size and shape of the surface. Proof: Let a point charge +q be placed at centre O of a sphere S. Then S is a Gaussian surface. Electric field at any point on S is given by, $\text{E}=\frac{1}{4\pi\epsilon_0}.\frac{\text{q}}{\text{r}^2}$Image
The electric field and area element points radially outwards, so $\theta=0^\circ,$
Flux through area $\vec{\text{dS}}$ is,
$\text{d}\Phi=\vec{\text{E}}.\text{dS}=\text{E dS}\cos0^\circ=\text{E dS}$ Total flux through surface S is, $\Phi=\oint\limits_{\text{S}}\text{d}\Phi=\oint\limits_{\text{S}}\text{E dS}=\text{E}\oint\limits_{\text{S}}\text{dS}$ = E × Area of Sphere $=\frac{1}{4\pi\epsilon_0}.\frac{\text{q}}{\text{x}^2}.4\pi\text{r}^2$ or, $\Phi=\frac{\text{q}}{\epsilon_0}$ which proves Gauss's theorem.
The electric field and area element points radially outwards, so $\theta=0^\circ,$
Flux through area $\vec{\text{dS}}$ is,
$\text{d}\Phi=\vec{\text{E}}.\text{dS}=\text{E dS}\cos0^\circ=\text{E dS}$ Total flux through surface S is, $\Phi=\oint\limits_{\text{S}}\text{d}\Phi=\oint\limits_{\text{S}}\text{E dS}=\text{E}\oint\limits_{\text{S}}\text{dS}$ = E × Area of Sphere $=\frac{1}{4\pi\epsilon_0}.\frac{\text{q}}{\text{x}^2}.4\pi\text{r}^2$ or, $\Phi=\frac{\text{q}}{\epsilon_0}$ which proves Gauss's theorem.
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