Gauss' Law

Which law is used to derive the expression for the electric field between two uniformly charged large parallel sheets with surface charge densities $\sigma$ and $-\sigma$ respectively:

A closed spherical surface encloses a charge $q$ at its centre. Show that the electric flux through the closed surface is $\frac{q}{{\epsilon }_0}$.

The inward and outward electric flux from a closed surface are respectively $8\times {10}^3$ and $4\times {10}^3$ units. Then the net charge inside the close surface is

An electric field $\mathrm{E}$ is present ${\perp }^{\mathrm{r}}$ to a square plate of side length $\mathrm{L}$. What is the flux passing through the plate?

A charge $\mathrm{Q}$ is distributed uniformly in a sphere (solid). Then the electric field at any point $\mathrm{r}$, where $\mathrm{r}<\mathrm{R}$ ( $\mathrm{r}$ is radius of sphere) varies as_____.

Two infinitely long parallel conducting plates having surface charge densities $+\mathrm{\sigma }$ and $-\mathrm{\sigma }$ respectively are separated by a small distance. The medium between the plates is vacuum, if ${\mathrm{\epsilon }}_{\mathrm{o}}$ is dielectric permittivity of vacuum, then the electric field in the region between the plates is-

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

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 $x$.

State Gauss' theorem in electrostatics and write its expression.