The electric flux for Gaussian surface A that enclose the charged particles in free space is (given $q_1$ = $-14\, nC$, $q_2$ = $78.85\, nC$, $q_3$ = $-56 \,nC$)

A positive charge $q$ is kept at the center of a thick shell of inner radius $R_1$ and outer radius $R_2$ which is made up of conducting material. If $\phi_1$ is flux through closed gaussian surface $S_1$ whose radius is just less than $R_1$ and $\phi_2$ is flux through closed gaussian surface $S_2$ whose radius is just greater than $R_1$ then:-

An electric field $\overrightarrow{\mathrm{E}}=(2 \mathrm{xi}) \mathrm{NC}^{-1}$ exists in space. $\mathrm{A}$ cube of side $2 \mathrm{~m}$ is placed in the space as per figure given below. The electric flux through the cube is .................. $\mathrm{Nm}^2 / \mathrm{C}$

If a charge $q$ is placed at the centre of a closed hemispherical non-conducting surface, the total flux passing through the flat surface would be

A point charge $+10\; \mu \,C$ is a distance $5 cm$ directly above the centre of a square of side $10 \;cm ,$ as shown in Figure. What is the magnitude of the electric flux through the square?

The circular wire in figure below encircles solenoid in which the magnetic flux is increasing at a constant rate out of the plane of the page. The clockwise emf around the circular loop is $\varepsilon_{0}$. By definition a voltammeter measures the voltage difference between the two points given by $V_{b}-V_{a}=-\int \limits_{a}^{b} E \cdot d s$ We assume that $a$ and $b$ are infinitesimally close to each other. The values of $V_{b}-V_{a}$ along the path $1$ and $V_{a}-V_{b}$ along the path $2$ , respectively are