Gauss's Law

Assertion: Gauss's law can't be used to calculate electric field near an electric dipole.
Reason: Electric dipole don't have symmetrical charge distribution.

Gauss law is applicable only when there is a symmetric distribution of charge.

A sphere of radius $\mathrm{R}$ have volume charge density given as
$\rho \left(\mathrm{r}\right)=\mathrm{Kr}$ for $\mathrm{r}\le \mathrm{R}$

$=0$ for $\mathrm{r}>\mathrm{R}$
If electric field at distance $\frac{\mathrm{R}}{2}$ from centre is $\frac{{\mathrm{KR}}^2}{2\mathrm{N}{\epsilon }_0}$ Then $\mathrm{N}$ will be.

The total flux through the faces of the cube with side of length $\mathrm{\alpha }$ if a charge $\mathrm{q}$ is placed at corner $\mathrm{A}$ of the cube is

A point charge of $1.8 \mathrm{\mu C}$ is at the centre of cubical Gaussian surface$55 \mathrm{cm}$ on edge. What is the net electric flux through the surface?

A ring of radius $R$ having a linear charge density $\lambda$ moves towards a solid imaginary sphere of radius $\frac{R}{2},$ so that the centre of ring passes through the centre of the sphere. The axis of the ring is perpendicular to the line joining the centres of the ring and the sphere. The maximum flux through the sphere in this process is

Flux coming out from a unit positive charge enclosed in air is

A Gaussian sphere encloses an electric dipole within it. The total flux across the sphere is

$q_1,\mathrm{ }{q}_2,\mathrm{ }{q}_3$ and $q_4$ are point charges located at points, as shown in the figure, and $S$ is a spherical Gaussian surface of radius $R$. Which of the following is true, according to the Gauss' law?

The figure shows a closed surface which intersects a conducting sphere. If a positive charge is placed at the point $P$, the flux of the electric field through the closed surface, 

A Gaussian sphere encloses an electric dipole within it. The total flux through the sphere is

Flux coming out from a unit positive charge enclosed in air is

The potential due to an electrostatic charge distribution is $V\left(r\right)=\frac{q{e}^{-\alpha r}}{4\pi {\epsilon }_{0}r},$ where $\alpha$ is positive. The net charge within a sphere centred at the origin and of radius $1/\alpha$ is

The electric field intensity outside the charged conducting sphere of radius $‘R’$, placed in a medium of permittivity $\epsilon$ at a distance $‘r’$ from the centre of the sphere in terms of surface charge density $\sigma$ is

A metal sphere of radius $1 \mathrm{cm}$ is charged with $3.14 \mathrm{\mu C}$. Find the electric intensity at a distance of $1 \mathrm{m}$ from the centre of the metal sphere.

A sphere of radius $\mathrm{R}$ have volume charge density given as
$\rho \left(\mathrm{r}\right)=\mathrm{Kr}$ for $\mathrm{r}\le \mathrm{R}$

$=0$ for $\mathrm{r}>\mathrm{R}$
If electric field at distance $\frac{\mathrm{R}}{2}$ from centre is $\frac{{\mathrm{KR}}^2}{\mathrm{N}{\epsilon }_0}$ Then $\mathrm{N}$ will be.

An uncharged thick spherical conducting shell is surrounding a charge $-q$ at the centre of the shell. Then charge $+3q$ is placed on a point outside the shell. When static equilibrium is reached, the total charges on the inner and outer surfaces of the shell are respectively

The charge density inside a nucleus is given by $\rho ={\rho }_0\left(1-\frac{r^2}{a^2}\right)$ where $r\le a,a=$radius of nucleus $=3 \mathrm{fm}$ and ${\rho }_0=5\times {10}^{25} \mathrm{C} {\mathrm{m}}^{-3}.$ The electric field is maximum at $r=r_0.$ If $r_0=\sqrt{\frac{5}{x}}a$, find the value of $x$.

Two conducting spheres of radii $r_1$ and $r_2$ have equal surface charge densities. The ratio of their charges is ___.

The radii of two conducting spheres are $a$ and $b$. When these are given same surface charge density the ratio of electric field intensities at their surfaces is