Obtain Gauss’s law from Coulomb’s law.
Obtain Gauss’s law from Coulomb’s law.
Coulombian force acting between charges $Q+q$ is,
$\mathrm{F}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{\mathrm{Q} q}{r^{2}}$
$\frac{\mathrm{F}}{\mathrm{Q}}=\frac{q}{4 \pi \varepsilon_{0} \cdot r^{2}}$
$\text { But, } \frac{\mathrm{F}}{\mathrm{Q}}=\overrightarrow{\mathrm{E}}$
[Force acting on Q charge placed in electric field of $q$ means intensity of electric field E.]
$\therefore \mathrm{E}=\frac{1}{4 \pi \varepsilon_{0}} \cdot \frac{q}{r^{2}}$
$\therefore \mathrm{E} \times 4 \pi r^{2}=\frac{q}{\varepsilon_{0}}$
$\therefore \int \mathrm{E} d s=\frac{q}{\varepsilon_{0}}, \text { where } 4 \pi r^{2}=d s$
As $\mathrm{E}$ and $d s$ are vectors,
$\int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d s}=\frac{q}{\varepsilon_{0}}$ This is Gauss's law.
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