The distance between the two charges $25\,\mu C$ and $36\,\mu C$ is $11\,cm$ At what point on the line joining the two, the intensity will be zero
At a distance of $5\,cm$ from $25\,\mu C$
At a distance of $5\,cm$ from $36\,\mu C$
At a distance of $10\,cm$ from $25\,\mu C$
At a distance of $11\,cm$ from $36\,\mu C$
A charged particle is suspended in equilibrium in a uniform vertical electric field of intensity $20000\, V/m$. If mass of the particle is $9.6 \times {10^{ - 16}}\,kg$, the charge on it and excess number of electrons on the particle are respectively $(g = 10\,m/{s^2})$
Give physical meaning of electric field.
A charged cork of mass $m$ suspended by a light string is placed in uniform electric filed of strength $E= $$(\hat i + \hat j)$ $\times$ $10^5$ $NC^{-1}$ as shown in the fig. If in equilibrium position tension in the string is $\frac{{2mg}}{{(1 + \sqrt 3 )}}$ then angle $‘\alpha ’ $ with the vertical is
Two charges $+Q$ and $-2 Q$ are located at points $A$ and $B$ on a horizontal line as shown below.The electric field is zero at a point which is located at a finite distance
A thin semi-circular ring ofradius $r$ has a positive charge $q$ distributed uniformly over it. The net field $\vec E$ at the centre $O$ is