$A$ charge of magnitude $3e$ and mass $2m$ is moving in an electric field $\overrightarrow{E}$. The acceleration imparted to the charge is:

An electron enters an electric field with its velocity in the direction of the electric lines of force. Then:

Charge $Q$ is given a displacement $\vec{r} = a\hat{i} + b\hat{j}$ in an electric field $\vec{E} = E_1\hat{i} + E_2\hat{j}$. The work done is

$A$ particle of mass $M$ and charge $q$ is released from rest in a region of uniform electric field of magnitude $E$. After a time $t$,the distance travelled by the charge is $S$ and the kinetic energy attained by the particle is $T$. Then,the ratio $T/S$:

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

An electron enters an electric field of intensity $3200 \ V/m$ perpendicular to the field with a velocity of $4 \times 10^7 \ m/s$. If it travels a horizontal distance of $0.10 \ m$,then its deflection from its path is ........ $mm$.