If an $\alpha$-particle and a proton are accelerated from rest by a potential difference of $1 \text{ MV}$,then the ratio of their kinetic energy will be

$A$ particle of charge $1\ \mu C$ and mass $1\ g$ moving with a velocity of $4\ m/s$ is subjected to a uniform electric field of magnitude $300\ V/m$ for $10\ s$. Then its final speed cannot be.......$m/s$.

The electric field at a point is given by $\vec E = E_0 \hat i \, V/m$. $A$ particle of charge $+q_0$ moves from point $A(0, a)$ to $B(a, 0)$ along a circular path as shown in the figure. Find the work done by the electric field in this motion.

As shown in the figure below,a point charge $q$ moves from point $P$ to a point $S$ traversing a path $PQRS$ in a uniform electric field $\vec{E}$. The electric field is directed along a direction parallel to the $x$-axis. The coordinates of $P$,$Q$,$R$,and $S$ are $(a, b, 0)$,$(2a, 0, 0)$,$(a, -b, 0)$,and $(0, 0, 0)$ respectively. What is the work done by the field in the process?

$A$ 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 $(g = 10 \,ms^{-2})$

$A$ particle of mass $1 \, g$ and charge $-0.1 \, \mu C$ is projected from the ground with a velocity $10\sqrt{2} \, m/s$ at an angle of $45^o$ with the horizontal in a region having a uniform electric field $1 \, kV/cm$ in the horizontal direction. Acceleration due to gravity is $10 \, m/s^2$ in the vertical downward direction. Select the $INCORRECT$ statement.