$A$ charged bead is sliding freely through a string held vertically under tension. An electric field is applied parallel to the string so that the bead stays at rest at the middle of the string. If the electric field is switched off momentarily and switched on again,then

$A$ particle having mass $m$ and charge $q$ is at rest. On applying a uniform electric field $E$ on it,it starts moving. When this particle travels a distance $x$ in the direction of the force,its kinetic energy will be . . . . . . .

$A$ uniform electric field $E = (8m/e) \text{ V/m}$ is created between two parallel plates of length $1 \text{ m}$ as shown in the figure (where $m = \text{mass of electron}$ and $e = \text{charge of electron}$). An electron enters the field symmetrically between the plates with a speed of $2 \text{ m/s}$. The angle of the deviation $(\theta)$ of the path of the electron as it comes out of the field will be:

$A$ particle of charge $q$ and mass $m$ is projected from origin with an initial velocity $\vec{v} = (\frac{v_0}{\sqrt{2}}\hat{i} + \frac{v_0}{\sqrt{2}}\hat{j})$. There exists a uniform magnetic field $\vec{B} = B_0\hat{z}$ and a space varying electric field $\vec{E} = E_0 e^{-\lambda x}\hat{x}$ within the region $0 \leq x \leq L$. After travelling a distance such that $x$-coordinate has changed from $x = 0$ to $x = L$, the change in the kinetic energy is . . . . . . .

An electron is rotating around an infinite positive linear charge in a circle of radius $0.1 \,m$. If the linear charge density is $1 \,\mu C/m$,then the velocity of the electron in $m/s$ will be ...... $\times 10^7$.

Two charges $-q$ each are fixed and separated by a distance $2d$. $A$ third charge $q$ of mass $m$ placed at the midpoint is displaced slightly by $x$ $(x \ll d)$ perpendicular to the line joining the two fixed charges as shown in the figure. Show that the charge $q$ will perform simple harmonic motion with a time period $T = \left[\frac{8 \pi^{3} \epsilon_{0} m d^{3}}{q^{2}}\right]^{1 / 2}$.