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
$\frac{q}{{4\pi {\varepsilon _0}{a^2}}}$
$\frac{{\sqrt 2 \,q}}{{4\pi {\varepsilon _0}{a^2}}}$
$\frac{{\sqrt 3 \,q}}{{4\pi {\varepsilon _0}{a^2}}}$
$\frac{q}{{2\pi {\varepsilon _0}{a^2}}}$
Two point charges $a$ and $b$, whose magnitudes are same are positioned at a certain distance from each other with a at origin. Graph is drawn between electric field strength at points between $a$ and $b$ and distance $x$ from a $E$ is taken positive if it is along the line joining from to be. From the graph, it can be decided that
The intensity of the electric field required to keep a water drop of radius ${10^{ - 5}}\, cm$ just suspended in air when charged with one electron is approximately
For the given figure the direction of electric field at $A$ will be
A total charge $q$ is divided as $q_1$ and $q_2$ which are kept at two of the vertices of an equilateral triangle of side a. The magnitude of the electric field $E$ at the third vertex of the triangle is to be depicted schematically as a function of $x=q_1 / q$. Choose the correct figure.
The magnitude of electric field intensity $E$ is such that, an electron placed in it would experience an electrical force equal to its weight is given by