If electric field in a region is given as $\overrightarrow{E}=\frac{E_0.x}{l}\hat{i}+\frac{E_0.y^2}{l^2}\hat{j}+\frac{E_0.z^3}{l^3}\hat{k} \left(E_0=5\times {10}^3 \frac{\mathrm{N}}{\mathrm{C}}, l=2 \mathrm{cm}\right)$, then calculate the flux of electric field passing through a rectangular plane $\left(2 \mathrm{cm}\times 1 \mathrm{cm}\right)$ at $y=2 \mathrm{cm}$

The electric flux through ring shown in figure 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 ${\mu }_s=0.5$)

Two balls of charges $q_1$ and $q_2$ initially have exactly same velocity. Both the balls are, subjected to same uniform electric field for same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by $60°.$ The direction of the velocity of second ball is found to change by $90°.$

The electric field and initial velocity of the charged particle are inclined at angle

Two balls of charges $q_1$ and $q_2$ initially have exactly same velocity. Both the balls are, subjected to same uniform electric field for same time. As a result, the velocity of the first ball is reduced to half of its initial value and its direction changes by $60°.$ The direction of the velocity of second ball is found to change by $90°.$

If new velocity of second charged particle has a magnitude $\text{'}x\text{'}$ times the initial velocity, then $x$ is