Obtain Coulomb’s law from Gauss’s law.
Obtain Coulomb’s law from Gauss’s law.
As shown in figure, consider a point charge $+q$ kept at $\mathrm{O}$.
A Gaussian surface ' $\mathrm{S}$ ' is shown in figure includes charge $q$.
Consider surface area $\overrightarrow{d s}$ at point $\mathrm{P}$. Here, $\overrightarrow{\mathrm{E}} \| \overrightarrow{d s}$ and so that $\theta=0^{\circ}$.
According to Gauss's law,
$\phi=\frac{q}{\varepsilon_{0}}$ $\therefore \int \overrightarrow{\mathrm{E}} \cdot \overrightarrow{d s}=\frac{q}{\varepsilon_{0}}$ $\therefore \int \mathrm{E} \cdot d s \cos 0^{\circ}=\frac{q}{\varepsilon_{0}}[\because \overrightarrow{\mathrm{E}} \| \overrightarrow{d s}]$ $\therefore \mathrm{E} \int d s=\frac{q}{\varepsilon_{0}}\left[\therefore \cos 0^{\circ}=1\right]$ $\therefore \mathrm{E} \times 4 \pi r^{2}=\frac{q}{\varepsilon_{0}}\left[\therefore \int d s=4 \pi r^{2}\right]$ $\therefore \mathrm{E}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}$ $\therefore \frac{\mathrm{F}}{q}=\frac{q}{4 \pi \varepsilon_{0} r^{2}}$ $\therefore \mathrm{F}=\frac{\mathrm{Kq} q_{0}}{r^{2}}$ $\mathrm{This}$ is Coulomb's law.
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List-$II$
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| List-$I$ | List-$II$ |
| $E$ is independent of $d$ | A point charge $Q$ at the origin |
| $E \propto \frac{1}{d}$ | A small dipole with point charges $Q$ at $(0,0, l)$ and $- Q$ at $(0,0,-l)$. Take $2 l \ll d$. |
| $E \propto \frac{1}{d^2}$ | An infinite line charge coincident with the x-axis, with uniform linear charge density $\lambda$ |
| $E \propto \frac{1}{d^3}$ | Two infinite wires carrying uniform linear charge density parallel to the $x$-axis. The one along ( $y=0$, $z =l$ ) has a charge density $+\lambda$ and the one along $( y =0, z =-l)$ has a charge density $-\lambda$. Take $2 l \ll d$ |
| plane with uniform surface charge density |
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