Electric field intensity due to point charge by Gauss's Law

Derivation of electric field intensity due to a point charge by Gauss's Law: Let us consider, a source point charge particle of $+q$ coulomb is placed at point $O$ in space. Let's take a point $P$ on the electric field of the source point charge particle. To find the electric field intensity $\overrightarrow{E}$ at point $P$, first put the test charge particle $+q_{0}$ on the point $P$ and draw a gaussian surface which passes through the point $P$. After that take a very small area $\overrightarrow{dA}$ around the point $P$. If the distance between the source charge particle and small area $\overrightarrow {dA}$ is $r$ then electric flux passing through the small area $\overrightarrow{dA}$ →

$ d\phi_{E}= \overrightarrow{E} \cdot \overrightarrow{dA}$

$ d\phi_{E}= E\:dA\: cos\theta$
Electric field due to point charge particle
Electric field due to point charge

from the figure, the direction between $\overrightarrow{E}$ and $\overrightarrow{dA}$ is parallel to each other i.e. the angle will be $0^{\circ}$. So the above equation can be written as →

$ d\phi_{E}= E\:dA\: cos0^{\circ}$

$ d\phi_{E}= E\:dA $

The electric flux passing through the entire Gaussian surface and be found by closed integration of the above equation →

$ d\phi_{E}= \oint {E\:dA} $

$ d\phi_{E}= E\:\oint {dA} $

$ \phi_{E}= E\left(4\pi r^{2} \right) \qquad \left\{ \because \oint {dA}=4\pi r^{2} \right\}$

According to Gauss's Law → $\phi_{E}= \frac{q}{\epsilon_{0}}$ then above equation can be written as →

$ \frac{q}{\epsilon_{0}}= E \left(4\pi r^{2} \right)$

$ E= \frac{1}{4\pi\epsilon_{0}} \frac{q}{r^{2}}$

The above expression is the electric field intensity due to a point source charged particle.

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