Two point charges $q_1\,(\sqrt {10}\,\,\mu C)$ and $q_2\,(-25\,\,\mu C)$ are placed on the $x-$ axis at $x = 1\,m$ and $x = 4\,m$ respectively. The electric field (in $V/m$ ) at a point $y = 3\,m$ on $y-$ axis is, [ take ${\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}$ ]
$(63\hat i - 27\hat j) \times {10^2}$
$(-63\hat i + 27\hat j) \times {10^2}$
$(81\hat i - 81\hat j) \times {10^2}$
$(-81\hat i + 81\hat j) \times {10^2}$
A charged particle of mass $0.003\, gm$ is held stationary in space by placing it in a downward direction of electric field of $6 \times {10^4}\,N/C$. Then the magnitude of the charge is
The direction $(\theta ) $ of $\vec E$ at point $P$ due to uniformly charged finite rod will be
Four charges $q, 2q, -4q$ and $2q$ are placed in order at the four corners of a square of side $b$. The net field at the centre of the square is
An infinite number of electric charges each equal to $5\, nC$ (magnitude) are placed along $X$-axis at $x = 1$ $cm$, $x = 2$ $cm$ , $x = 4$ $cm$ $x = 8$ $cm$ ………. and so on. In the setup if the consecutive charges have opposite sign, then the electric field in Newton/Coulomb at $x = 0$ is $\left( {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,N - {m^2}/{c^2}} \right)$
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 ${O}$ ?