Four charges are placed on corners of a square as shown in figure having side of $5\,cm$. If $Q$ is one microcoulomb, then electric field intensity at centre will be

If the net electric field at point $\mathrm{P}$ along $\mathrm{Y}$ axis is zero, then the ratio of $\left|\frac{q_2}{q_3}\right|$ is $\frac{8}{5 \sqrt{x}}$, where $\mathrm{x}=$. . . . . .

Two point charges $q_1$ and $q_2 (=q_1/2)$ are placed at points $A(0, 1)$ and $B(1, 0)$ as shown in the figure. The electric field vector at point $P(1, 1)$ makes an angle $\theta $ with the $x-$ axis, then the angle $\theta$ is

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

A wire of length $L\, (=20\, cm)$, is bent into a semicircular arc. If the two equal halves of the arc were each to be uniformly charged with charges $ \pm Q\,,\,\left[ {\left| Q \right| = {{10}^3}{\varepsilon _0}} \right]$ Coulomb where $\varepsilon _0$ is the permittivity (in $SI\, units$) of free space] the net electric field at the centre $O$ of the semicircular arc would be

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