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
$x = \frac{a}{{\sqrt 2 }}$
$x = \sqrt 2 a$
$x = \frac{{\sqrt 2 a}}{{\sqrt 2 - 1}}$
$x = \frac{{\sqrt 2 a}}{{\sqrt 2 + 1}}$
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
Two point charges $q_{1}$ and $q_{2},$ of magnitude $+10^{-8} \;C$ and $-10^{-8}\; C ,$ respectively, are placed $0.1 \;m$ apart. Calculate the electric fields at points $A, B$ and $C$ shown in Figure
Two point charges of $20\,\mu \,C$ and $80\,\mu \,C$ are $10\,cm$ apart. Where will the electric field strength be zero on the line joining the charges from $20\,\mu \,C$ charge......$m$
Five charges, $\mathrm{q}$ each are placed at the corners of a regular pentagon of side $\mathrm{'a'}$ as in figure.
$(a)$ $(i)$ What will be the electric field at $O$, the centre of the pentagon ?
$(ii)$ What will be the electric field at $O$ if the charge from one of the corners (say $A$ $)$ is removed ?
$(iii)$ What will be the electric field at $O $ if the charge $q$ at $A$ is replaced by$ -q$ ?
$(b) $ How would your answer to $(a)$ be affected if pentagon is replaced by $n\,-$ sided regular polygon with charge $q$ at each of its corners ?
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 ${O}$ ?