Two identical charges $+Q$ are kept fixed at some distance apart. $A$ small particle $P$ with charge $q$ is placed midway between them. If $P$ is given a small displacement $\Delta$,it will undergo simple harmonic motion if:

$A$ negative point charge is placed at point $A$ as shown in the figure. The charge is:

Four equal positive charges are fixed at the vertices of a square of side $L$. The $Z$-axis is perpendicular to the plane of the square. The point $z = 0$ is the point where the diagonals of the square intersect. Find the plot of the electric field $E$ due to the four charges as one moves along the $Z$-axis.

In the given figure,two tiny conducting balls of identical mass $m$ and identical charge $q$ hang from non-conducting threads of equal length $L$. Assume that $\theta$ is so small that $\tan \theta \approx \sin \theta$,then for equilibrium,$x$ is equal to:

$A$ tiny spherical oil drop carrying a net charge $q$ is balanced in still air with a vertical uniform electric field of strength $\frac{81 \pi}{7} \times 10^5 \text{ Vm}^{-1}$. When the field is switched off,the drop is observed to fall with a terminal velocity $2 \times 10^{-3} \text{ ms}^{-1}$. Given $g = 9.8 \text{ ms}^{-2}$,viscosity of the air $\eta = 1.8 \times 10^{-5} \text{ Ns m}^{-2}$,and the density of oil $\rho = 900 \text{ kg m}^{-3}$,the magnitude of $q$ is:

$A$ Van de Graaff generator produces: