$A$ $3 \text{ C}$ charge moves from the point $(0, -2, -5)$ to the point $(5, 1, 2)$ in an electric field expressed as $\vec{E} = 2\hat{i} + 3\hat{j} + 4\hat{k} \text{ N/C}$. The work done in moving the charge is . . . . . . $J$.

$A$ particle of mass '$m$' and charge '$q$',initially at rest, is accelerated by a uniform electric field '$E$' through a distance '$D$' and is then allowed to approach a fixed static charge '$Q$' of the same sign. The distance of the closest approach of the charge '$q$' is [ $\epsilon_0$ = permittivity of free space ]

The velocity acquired by an electron at rest when subjected to a uniform electric field of potential difference $180 \ V$ is (Mass of electron $= 9 \times 10^{-31} \ kg$ and charge of electron $= 1.6 \times 10^{-19} \ C$) (in $km \ s^{-1}$)

$A$ negatively charged particle $p$ is placed,initially at rest,in a constant,uniform gravitational field and a constant,uniform electric field as shown in the diagram. What,qualitatively,is the shape of the trajectory of the particle?

Two charges each of magnitude $q$ are fixed at a distance $2d$ apart. $A$ third charge (proton) placed at the midpoint is displaced slightly by a distance $x$ $(x << d)$ perpendicular to the line joining the two fixed charges. The proton will execute simple harmonic motion having an angular frequency: ($m =$ mass of the charged particle)

As shown in the following figure,an electron falls through a distance of $1.5 \ cm$ in a uniform electric field of magnitude $2.0 \times 10^4 \ NC^{-1}$. Find the acceleration of the electron due to the electric field.