91Ó°ÊÓ

• The charges of the particles, q₁=q₂=+e, q₃=+2e.
• The distance, a=6mm=0.006 m

The electric field is a vector field, so the net electric field at any point can be calculated by doing the vector addition of all the fields due to the individual sources.

The electric field is given as,

$\stackrel{→}{\mathrm{E}}=\frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}{\mathrm{R}}^2}\hat{\mathrm{R}}$ (i)

where, R is the distance of field point from the charge and q is the charge on the particle

By symmetry we see that the contributions from the two charges q₁=q₂=+e cancel each other.

Thus, the magnitude of the net electric field is due to charge 3 which is located at a distance of $\frac{\mathrm{a}}{\sqrt{2}}$from P. the magnitude of net electric field $\left|\stackrel{→}{{\mathrm{E}}_{\mathrm{net}}}\right|$is given using equation (i) as:

$\begin{array}{rcl}\left|\stackrel{→}{{\mathrm{E}}_{\mathrm{net}}}\right|& =& \frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}\frac{2\mathrm{e}}{{\left({\displaystyle \frac{\mathrm{a}}{\sqrt{2}}}\right)}^2}\\ & =& \frac{1}{4{\mathrm{Ï€Î\mu }}_{\mathrm{o}}}\frac{4\mathrm{e}}{{\mathrm{a}}^2}\\ & =& \frac{\left(8.99×{10}^9\mathrm{N}.{\displaystyle \frac{{\mathrm{m}}^2}{{\mathrm{C}}^2}}\right)}{\left(6.0×{10}^{-6}{\mathrm{m}}^2\right)}\left[1.6×{10}^{-19}\mathrm{C}\right]\\ & =& 160\mathrm{N}/\mathrm{C}\end{array}$

Hence, the value of the net electric field is 160 N/C

As the net electric field must be directed away from the charge q₃ and the length of lines 2P and 3P are equal, so the angle $\left(\mathrm{θ}\right)$made by the net electric field with +x-axis is given as-

$\begin{array}{rcl}\mathrm{θ}& =& {\tan }^{-1}\left(\frac{2\mathrm{P}}{3\mathrm{P}}\right)\\ & =& {\tan }^{-1}\left(1\right)\\ & =& {45}^{\mathrm{o}}\end{array}$

Thus, the electric field points in the direction making an angle of 45^o with the positive side of the x-axis.