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Obtain Gauss’s law from Coulomb’s law.
Solution
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.
Similar Questions
The electric field $E$ is measured at a point $P (0,0, d )$ generated due to various charge distributions and the dependence of $E$ on $d$ is found to be different for different charge distributions. List-$I$ contains different relations between $E$ and $d$. List-$II$ describes different electric charge distributions, along with their locations. Match the functions in List-$I$ with the related charge distributions in List-$II$.
| 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 |
