Two charges $ + 5\,\mu C$ and $ + 10\,\mu C$ are placed $20\, cm$ apart. The net electric field at the mid-Point between the two charges is
$4.5 \times {10^6}\, N/C$ directed towards $ + 5\,\mu C$
$4.5 \times {10^6}\, N/C$ directed towards $ + 10\,\mu C$
$13.5 \times {10^6}\, N/C$ directed towards $ + 5\,\mu C$
$13.5 \times {10^6}\, N/C$ directed towards $ + 10\,\mu C$
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 acceleration of an electron in an electric field of magnitude $50\, V/cm$, if $e/m$ value of the electron is $1.76 \times {10^{11}}\,C/kg$, is
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......$m$
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 ${O}$ ?
Deutron and $\alpha - $ particle are put $1\,\mathop A\limits^o $ apart in air. Magnitude of intensity of electric field due to deutron at $\alpha - $ particle is