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 $\mathrm{x}=$. . . . . .
$2$
$3$
$5$
$6$
Is electric field scalar or vector ? Why ?
Whose result the whole electrostatic is ?
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 ${O}$ ?
The electric field intensity just sufficient to balance the earth's gravitational attraction on an electron will be: (given mass and charge of an electron respectively are $9.1 \times 10^{-31}\,kg$ and $1.6 \times$ $10^{-19}\,C$.)
Two point charges $Q_1, Q_2$ are fixed at $x = 0$ and $x = a$. Assuming that field strength is positive in the direction coinciding with the positive direction of $x$, then, which following option will be correct ?