If electric field in a region is given by
$\overrightarrow{E}=\frac{E_0.x}{l}\hat{i}+\frac{E_0.y^2}{l^2}\hat{j}+\frac{E_0.z^3}{l^3}\hat{k}$
where $E_0=5\times {10}^3 \frac{\mathrm{N}}{\mathrm{C}}, l=2 \mathrm{cm}.$
then electric flux passing through the plane $x=2 \mathrm{cm}.$ The plane is a square of side $2 \mathrm{cm}.$

The electric flux through ring shown in figure is

Charge is uniformly distributed with volume charge density $\rho$ in a spherical volume of radius $R.$ A cavity of radius $r$ is made in the charge distribution such that the centre of the cavity is at position vector $\overrightarrow{a}$ from the centre of the charge distribution, then the electric field in the cavity is

A dipole having charges $+q$ and $–q$ is fixed on a rough surface as shown. Another charge $Q$ of mass $m$ is placed at distance $d$ from this dipole. The minimum value of $d$ for which charge $Q$ will not move. (Given $a<<d$, co-efficient of static friction ${\mu }_s=0.5$)