Six charges, three positive and three negative of equal magnitude are to be placed at the vertices of a regular hexagon such that the electric field at $O$ is double the electric field when only one positive charge of same magnitude is placed at $R$. Which of the following arrangements of charges is possible for $P,\,Q,\,R,\,S,\,T,\,$ and $U$ respectively
$ + ,\, - ,\, + ,\, - ,\, - ,\, + $
$ + ,\, - ,\, + ,\, - ,\, + ,\, - $
$ + ,\, + ,\, - ,\, + ,\, - ,\, - $
$ - ,\, + ,\, + ,\, - ,\, + ,\, - $
Two point charges $A$ and $B$ of magnitude $+8 \times 10^{-6}\,C$ and $-8 \times 10^{-6}\,C$ respectively are placed at a distance $d$ apart. The electric field at the middle point $O$ between the charges is $6.4 \times 10^{4}\,NC ^{-1}$. The distance ' $d$ ' between the point charges $A$ and $B$ is..............$m$
Figure shows a rod ${AB}$, which is bent in a $120^{\circ}$ circular arc of radius $R$. A charge $(-Q)$ is uniformly distributed over rod ${AB}$. What is the electric field $\overrightarrow{{E}}$ at the centre of curvature ${O}$ ?
Four point charges $-q, +q, +q$ and $-q$ are placed on $y$ axis at $y = -2d$, $y = -d, y = +d$ and $y = +2d$, respectively. The magnitude of the electric field $E$ at a point on the $x -$ axis at $x = D$, with $D > > d$, will vary as
The electric field intensity at a point in vacuum is equal to
A drop of ${10^{ - 6}}\,kg$ water carries ${10^{ - 6}}\,C$ charge. What electric field should be applied to balance its weight (assume $g = 10\,m/{s^2}$)