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
${\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
${\tan ^{ - 1}}\left( {\frac{1}{4}} \right)$
${\tan ^{ - 1}}\left( 1 \right)$
${\tan ^{ - 1}}\left( 0 \right)$
A liquid drop having $6$ excess electrons is kept stationary under a uniform electric field of $25.5\, k\,Vm^{-1}$ . The density of liquid is $1.26\times10^3\, kg\, m^{-3}$ . The radius of the drop is (neglect buoyancy)
Mention characteristics of electric field.
The insulation property of air breaks down at $E = 3 \times {10^6}$ $volt\,/\,metre$. The maximum charge that can be given to a sphere of diameter $5\,m$ is approximately (in coulombs)
Charge $q$ is uniformly distributed over a thin half ring of radius $R$. The electric field at the centre of the ring is
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