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
$\frac{q}{\varepsilon_{0}}$
$\frac{q}{2 \varepsilon_{0}}$
$\frac{q}{4 \varepsilon_{0}}$
$\frac{ q }{2 \pi \varepsilon_{0}}$
Gauss’s law states that
A square surface of side $L$ metres is in the plane of the paper. A uniform electric field $\vec E(V/m) $, also in the plane of the paper, is limited only to the lower half of the square surface, (see figure). The electric flux in SI units associated with the surface is
Let the electrostatic field $E$ at distance $r$ from a point charge $q$ not be an inverse square but instead an inverse cubic, e.g. $E =k \cdot \frac{q}{r^{3}} \hat{ r }$, here $k$ is a constant.
Consider the following two statements:
$(I)$ Flux through a spherical surface enclosing the charge is $\phi=q_{\text {enclosed }} / \varepsilon_{0}$.
$(II)$ A charge placed inside uniformly charged shell will experience a force.
Which of the above statements are valid?
A hollow cylinder has a charge $q$ coulomb within it. If $\phi$ is the electric flux in units of $volt-meter$ associated with the curved surface $B,$ the flux linked with the plane surface $A$ in units of $V-m$ will be
Why do electric field lines not form closed loop ?