$A$ $3 \text{ C}$ 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 . . . . . . $J$.

$A$ moving charge will gain energy due to the application of

Two large vertical and parallel metal plates having a separation of $1 \ cm$ are connected to a $DC$ voltage source of potential difference $X$. $A$ proton is released at rest midway between the two plates. It is found to move at $45^{\circ}$ to the vertical $JUST$ after release. Then $X$ is nearly

$A$ charged particle with charge $q$ and mass $m$ starts with an initial kinetic energy $K$ at the center of a uniformly charged spherical region of total charge $Q$ and radius $R$. $q$ and $Q$ have opposite signs. The spherically charged region is not free to move. The value of $K$ is such that the particle will just reach the boundary of the spherically charged region. How much time does it take for the particle to reach the boundary of the region?

$A$ body having a specific charge of $8\,\mu \text{C/g}$ is resting on a frictionless plane at a distance of $10\,\text{cm}$ from a wall. It starts moving towards the wall when a uniform electric field of $100\,\text{V/m}$ is applied horizontally toward the wall. If the collision of the body with the wall is perfectly elastic,then the time period of the motion will be (in seconds):

An electron of mass $m$,charge $e$ falls through a distance $h$ meters in a uniform electric field $E$. Then the time of fall is: