Q.
Two point charges $Q_1 = 2 \mu C$ and $ Q_2 = 1 \mu C$ are placed as shown. The coordinates of the point P are (2 cm, 1 cm). The electric intensity vector at P subtends an angle $\theta$ with the positive X-axis. The value of $\theta$ is given by
Solution:
Here, $Q_1 = 2\mu C$
$ \, \, \, \, \, \, \, \, \, \, Q_2= 1 \mu C$
$r_1 =AP= 2\: cm = 0.02\: m$
$r_2 = BP = 1 \: cm = 0.01 \: m$
From diagram, $\tan \theta = \frac{E_2}{E_1} $
$=\frac{Q_{2}}{Q_{1} } \times\left(\frac{r_{1}}{r_{2}}\right)^{2} = \frac{1}{2} \times \left(\frac{2}{1}\right)^{2} = 2$
