In a certain region,a uniform electric field $E$ and a magnetic field $B$ are present in opposite directions. At the instant $t = 0$,a particle of mass $m$ carrying a charge $q$ is given a velocity $v_0$ at an angle $\theta$ with the $y$-axis,in the $yz$-plane. The time after which the speed of the particle would be minimum is equal to

$A$ particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = (\frac{v_0}{\sqrt{2}}\hat{i} + \frac{v_0}{\sqrt{2}}\hat{j})$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is . . . . . . .

The electric potential varies in space according to the relation $V = 3x + 4y$. $A$ particle of mass $0.1 \, kg$ starts from rest from point $(2, 3.2)$ under the influence of this field. The charge on the particle is $+1 \, \mu C$. Assume $V$ and $(x, y)$ are in $S.I.$ units. The time taken to cross the $x$-axis is.....$s$

$A$ charged particle,initially at rest at $O$,follows the trajectory shown alongside when released. Such a trajectory is possible in the presence of:

As shown in the figure, an electron is projected with kinetic energy $K$ at an angle of $\theta = 45^\circ$ between two charged plates. The electron strikes the upper plate. The value of the electric field $E$ must be greater than:

$A$ particle of mass $m$ and charge $q$ is released from rest in a uniform electric field. If there is no other force on the particle,the dependence of its speed $v$ on the distance $x$ travelled by it is correctly given by (graphs are schematic and not drawn to scale)