91Ó°ÊÓ

Two identical positive charges of magnitude$3.0nC$are separated by a distance of$5cm$from the center of separation between the charges.

We have to find the net electric field at point p. Let $\mathrm{Ï•}$be the angle between AP and AB.

From the figure, $AP=\sqrt{2cm}andBP=5\sqrt{2cm}$

electric field at P due to the charge at B is given by,

$E_1=9×{10}^9×\frac{3×{10}^{-9}}{{\left(5\sqrt{2}×{10}^{-2}\right)}^2}=5400\mathrm{N}/\mathrm{C}$

The direction of $\stackrel{→}{E1}$is along BP.

So electric field at P due to the charge at A is given by,

$E_2=9×{10}^9×\frac{3×{10}^{-9}}{{\left(5\sqrt{2}×{10}^{-2}\right)}^2}=5400\mathrm{N}/\mathrm{C}$

The direction of $\stackrel{→}{E2}$is along AP.

Now $\stackrel{→}{E_1}and\stackrel{→}{E_2}$is resolved into parallel and perpendicular components.

Since the perpendicular components are equal and opposite they cancel each other.

Therefore net electric field at P is given by,

$E=E_1\cos \mathrm{θ}+E_2\cos \mathrm{θ}=2E_1\cos \mathrm{θ}\text{since}{E}_1=E_2$

From the figure, $$\begin{gathered} \cos \mathrm{θ}=\frac{1}{\sqrt{2}} \\ E=2×5400×\frac{1}{\sqrt{2}}=7637\mathrm{N}/\mathrm{C} \end{gathered}$$

The direction of the net electric field at P is along the horizontal.

Since $\cos \mathrm{θ}=\frac{1}{\sqrt{2}},\mathrm{θ}={45}^0$

So the direction of $\stackrel{→}{E_1}is45°$above the horizontal and direction of$\stackrel{→}{E_2}is45°$below the horizontal.