Gauss’s Law

The SI unit of electrical flux is:

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:

Applying Gauss theorem, the expression for the electric field intensity at a point due to an infinitely long, thin, uniformly charged straight wire is

Assertion (A): In a region of constant potential, the electric field is zero and there can be no charge inside the region.

Reason (R): According to Gauss law. charge inside the region should be zero if electric field is zero.

A solid sphere of radius $R$ carries a positive charge $Q$ distributed uniformly throughout its volume. A very thin hole is drilled through it's center. A particle of mass $m$ and charge $-q$ performs simple harmonic motion about the center of the sphere in this hole. The frequency of oscillation is

A hollow charged metal sphere has radius $r$. If the potential difference between its surface and a point at a distance $3r$ from the centre is$V$, then electric field intensity at a distance $3r$ is:

A cylinder of radius $R$ and length $L$ is placed in a uniform electric field $E$ parallel to the cylinder axis. The total flux for the surface of the cylinder is given by

An $100\mathrm{eV}$ electron is fired directly towards a large metal plate having surface charge density $-2\times {10}^{-6}{ \mathrm{C} \mathrm{m}}^{-2}$. The distance from where the electron be projected so that it just fails to strike the plate is

A circular plate sheet of radius $10 \mathrm{cm}$ is placed in a uniform electric field of $2\sqrt{3}\times {10}^5 \mathrm{N} {\mathrm{C}}^{-1},$ making an angle of $60°$ with the field. Then find the electric flux through the sheet.

A hollow cylinder has a charge $q$ coulomb within it. If $\varphi$ is the electric flux in units of $\mathrm{Volt} \mathrm{meter}$ associated with the curved surface $\mathrm{B}$, the flux linked with the plane surface $\mathrm{A}$ in units of $\mathrm{Volt} \mathrm{meter}$ will be

The $\mathrm{ }\mathrm{S}\mathrm{I}$ unit of electric flux is:

A parallel plate capacitor has two square plates with equal and opposite charges. The surface charge densities on the plates are $+\mathrm{\sigma }$ and $–\mathrm{\sigma }$ respectively. In the region between the plates the magnitude of the electric field is

The potential (in volts) of a charge distribution is given by

$V\left(z\right)=30-5z^2$ for $\left|z\right|\le 1\mathrm{ }\mathrm{m}$

$V\left(z\right)=35-10\mathrm{ }\left|\mathrm{z}\right|$ for $\left|Z\right|\ge 1\mathrm{ }\mathrm{m}$ .

$V\left(z\right)$ does not depend on $x$ and $y$. If this potential is generated by a constant charge per unit volume ${\mathrm{\rho }}_0$ (in units of ${\mathrm{ϵ}}_0$ ) which is spread over a certain region, then choose the correct statement.

The $\mathrm{ }\mathrm{S}\mathrm{I}$ unit of electric flux is:

The ratio of electric field intensity at $\mathrm{P}$ & $\mathrm{Q}$ in the shown arrangement is

A charge $Q$ is placed at a distance $\raisebox{1ex}{$a$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ above the centre of a square surface of side length $a$. The electric flux through the square surface due to the charge would be?

The net flux passing through the surface of an imaginary cube of edge length $\mathrm{a}$ placed in space having a uniform volume charge density, is $\mathrm{\varphi }$. The net flux passing through the surface of an imaginary sphere of radius $\mathrm{a}$ in the same space is

Consider a uniformly charged solid cylinder of large length and radius $R.$ Now consider a cylindrical surface of radius $2R$ and length $2R$ coaxial with cylinder and electric flux through cylindrical surface is ${\varphi }_0.$ Now consider a spherical surface of radius $2R$ and one of the diameter along axis of cylinder and electric flux through spherical surface is $\varphi .$ Find $10\frac{\varphi }{{\varphi }_0}$ to the nearest integer.

In the shown figure a hemisphere is placed and two charges $3\mathrm{q}$ and $9\mathrm{q}$ are placed symmetrically about centre $O.$ Net electric flux over curved surface is given as $\frac{N_{1}q}{N_2{\epsilon }_0},$ where $N_1$ and $N_2$ are lowest possible integers then value of $N_1+N_2$ is :