A drop of ${10^{ - 6}}\,kg$ water carries ${10^{ - 6}}\,C$ charge. What electric field should be applied to balance its weight (assume $g = 10\,m/{s^2}$)

Two point charge $-q$ and $+q/2$ are situated at the origin and at the point $(a, 0, 0)$ respectively. The point along the $X$ - axis where the electric field vanishes is

Five point charge each having magnitude $‘q’$ are placed at the corner of hexagon as shown in fig. Net electric field at the centre $‘O’$ is $\vec E$. To get net electric field at $‘O’$ be $6\vec E$, charge placed on the remaining sixth corner should be

Equal charges $q$ are placed at the vertices $A$ and $B$ of an equilateral triangle $ABC$ of side $a$. The magnitude of electric field at the point $C$ is

The tiny ball at the end of the thread shown in 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 .....$\mu C$

Find out electric filed intensity at point $A(1,0,2)$ due to a point charge $-20\,\mu C$ situated at point $B(0, 2,1)$