For a closed surface $\oint {\overrightarrow {E \cdot } } \,\overrightarrow {ds} \,\, = \,\,0$, then
Electric field at every point on surface is zero
Electric field at every point on surface is uniform
Electric Field at every point on surface is parallel
The number of electric field lines entering the surface will be equal to number of electric field line exit the surface
A point charge $+Q$ is positioned at the centre of the base of a square pyramid as shown. The flux through one of the four identical upper faces of the pyramid is
An infinite, uniformly charged sheet with surface charge density $\sigma$ cuts through a spherical Gaussian surface of radius $R$ at a distance $x$ from its center, as shown in the figure. The electric flux $\Phi $ through the Gaussian surface is
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
Give reason : ''If net flux assocaited with closed surface is zero, then net charge enclosed by that surface is zero''.
Draw electric field lines when two positive charges are near.