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.
Similar Questions
Obtain the expression of electric field by charged spherical shell on a point outside it.
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A conducting sphere of radius $R = 20$ $cm$ is given a charge $Q = 16\,\mu C$. What is $\overrightarrow E $ at centre
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A spherically symmetric charge distribution is considered with charge density varying as
$\rho(r)=\left\{\begin{array}{ll}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{array}\right.$
Where, $r ( r < R )$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be.
A spherically symmetric charge distribution is considered with charge density varying as
$\rho(r)=\left\{\begin{array}{ll}\rho_{0}\left(\frac{3}{4}-\frac{r}{R}\right) & \text { for } r \leq R \\ \text { Zero } & \text { for } r>R\end{array}\right.$
Where, $r ( r < R )$ is the distance from the centre $O$ (as shown in figure). The electric field at point $P$ will be.
- [JEE MAIN 2022]
Two non-conducting solid spheres of radii $R$ and $2 \ R$, having uniform volume charge densities $\rho_1$ and $\rho_2$ respectively, touch each other. The net electric field at a distance $2 \ R$ from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio $\frac{\rho_1}{\rho_2}$ can be ;
$(A)$ $-4$ $(B)$ $-\frac{32}{25}$ $(C)$ $\frac{32}{25}$ $(D)$ $4$
Two non-conducting solid spheres of radii $R$ and $2 \ R$, having uniform volume charge densities $\rho_1$ and $\rho_2$ respectively, touch each other. The net electric field at a distance $2 \ R$ from the centre of the smaller sphere, along the line joining the centres of the spheres, is zero. The ratio $\frac{\rho_1}{\rho_2}$ can be ;
$(A)$ $-4$ $(B)$ $-\frac{32}{25}$ $(C)$ $\frac{32}{25}$ $(D)$ $4$
- [IIT 2013]
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is $\left(\sigma / 2 \varepsilon_{0}\right) \hat{ n },$ where $\hat{ n }$ is the unit vector in the outward normal direction, and $\sigma$ is the surface charge density near the hole.