Two charged particles of masses in the ratio $1: 3$ have charges in reciprocal ratio as their masses. They are placed in a uniform electric field and allowed to move. The ratio of their kinetic energies is

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

$A$ water drop of mass $10^{-6} \ kg$ carries a charge of $(-10^{-6}) \ C$. What magnitude of electric field should be applied to the drop so that it is in equilibrium with its weight?

$A$ small oil drop of mass $10^{-6} \,kg$ is hanging at rest between two plates separated by $1 \,mm$ having a potential difference of $500 \,V$. The charge on the drop is $(g = 10 \,ms^{-2})$

$A$ mass $m = 20\,g$ has a charge $q = 3.0\,mC$. It moves with a velocity of $20\,m/s$ and enters a region of electric field of $80\,N/C$ in the same direction as the velocity of the mass. The velocity of the mass after $3\,s$ in this region is.......$m/s$.

In a region of space,suppose there exists a uniform electric field $\vec{E} = 10 \hat{i} \text{ V/m}$. If a positive charge moves with a velocity $\vec{v} = -2 \hat{j} \text{ m/s}$,its potential energy: