The tiny ball at the end of the thread shown in the figure has a mass of $0.5 \, g$ and is placed in a horizontal electric field of intensity $500 \, N/C$. It is in equilibrium in the position shown. The magnitude and sign of the charge on the ball is .....$\mu C$.

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)

$A$ particle of mass $m$ and charge $(-q)$ enters the region between two charged plates,initially moving along the $x$-axis with a speed $v_{x} = 2.0 \times 10^{6} \; m \, s^{-1}$. If the electric field $E$ between the plates,which are separated by $0.5 \; cm$,is $9.1 \times 10^{2} \; N/C$,at what distance along the $x$-axis will the electron strike the upper plate (in $cm$)?
$(|e| = 1.6 \times 10^{-19} \; C, m_{e} = 9.1 \times 10^{-31} \; kg)$

Select the correct statement: (Only force on a particle is due to electric field)

$A$ vertical electric field of magnitude $4.9 \times 10^{5} \, N/C$ just prevents a water droplet of mass $0.1 \, g$ from falling. The value of charge on the droplet will be ........ $\times 10^{-9} \, C$ (Given $g = 9.8 \, m/s^{2}$)

An electron having charge '$e$' and mass '$m$' is moving in a uniform electric field $E$. Its acceleration will be