A pendulum bob of mass $30.7 \times {10^{ - 6}}\,kg$ and carrying a charge $2 \times {10^{ - 8}}\,C$ is at rest in a horizontal uniform electric field of $20000\, V/m$. The tension in the thread of the pendulum is $(g = 9.8\,m/{s^2})$
$3 \times {10^{ - 4}}\,N$
$4 \times {10^{ - 4}}\,N$
$5 \times {10^{ - 4}}\,N$
$6 \times {10^{ - 4}}\,N$
Three charges are placed as shown in figure. The magnitude of $q_1$ is $2.00\, \mu C$, but its sign and the value of the charge $q_2$ are not known. Charge $q_3$ is $+4.00\, \mu C$, and the net force on $q_3$ is entirely in the negative $x-$ direction. The magnitude of $q_2$ is
The electric field due to a charge at a distance of $3\, m$ from it is $500\, N/coulomb$. The magnitude of the charge is.......$\mu C$ $\left[ {\frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {{10}^9}\,\frac{{N - {m^2}}}{{coulom{b^2}}}} \right]$
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
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 ${\mkern 1mu} {\mkern 1mu} \frac{1}{{4\pi {\varepsilon _0}}} = 9 \times {10^9}{\mkern 1mu} {\mkern 1mu} N{m^2}{C^{ - 2}}{\rm{ }}$ ]
Charges $Q _{1}$ and $Q _{2}$ arc at points $A$ and $B$ of a right angle triangle $OAB$ (see figure). The resultant electric field at point $O$ is perpendicular to the hypotenuse, then $Q _{1} / Q _{2}$ is proportional to