$A$ 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)

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?

An electron is made to enter symmetrically between two parallel and equally but oppositely charged metal plates,each of $10 \ cm$ length. The electron emerges out of the field region with a horizontal component of velocity $10^6 \ m/s$. If the magnitude of the electric field between the plates is $9.1 \ V/cm$,then the vertical component of velocity of the electron is (mass of electron $= 9.1 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$)

An oil drop having charge $2e$ is kept stationary between two parallel horizontal plates $2.0 \, cm$ apart when a potential difference of $12000 \, V$ is applied between them. If the density of oil is $900 \, kg/m^3$,the radius of the drop will be

Two masses $M_1$ and $M_2$ carry positive charges $Q_1$ and $Q_2$,respectively. They are dropped to the floor in a laboratory setup from the same height,where there is a constant electric field vertically upwards. $M_1$ hits the floor before $M_2$. Then,

There is a uniform electric field of strength $10^3 \ Vm^{-1}$ along the $Y$-axis. $A$ body of mass $1 \ g$ and charge $10^{-6} \ C$ is projected into the field from the origin along the positive $X$-axis with a velocity of $10 \ ms^{-1}$. Its speed in $ms^{-1}$ after $10 \ s$ is (Neglect gravitation).