Question

What are the strength and direction of the electric field at the position indicated by the dot in FIGURE EX23.2? Specify the direction as an angle above or below horizontal.

Short answer

The strength of the electric field at P$=1.33\times {10}^{4}N/C$

The direction of the net electric field at P is$18°$above the horizontal.

Step-by-step solution

Given information

Two positive charges of magnitude$3.0nCand6.0nC$are separated by a distance of$10cm$. The dot is at a perpendicular distance of$5cm$from charge$3.0nC$.

Explanation

We have to find the net electric field at the point P. Let$\varphi$be the angle between AP and PB.

From the figure, $\mathrm{BP}=5\sqrt{5}\mathrm{cm}$

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

$E_1=9\times {10}^9\times \frac{3\times {10}^{-9}}{{\left(5\times {10}^{-2}\right)}^2}=1.08\times {10}^4\mathrm{N}/\mathrm{C}$

The direction of $\overrightarrow{E_1}$is along the horizontal.

$E_2=9\times {10}^9\times \frac{6\times {10}^{-9}}{{\left(5\sqrt{5}\times {10}^{-2}\right)}^2}=0.432\times {10}^4\mathrm{N}/\mathrm{C}$

The direction of $\overrightarrow{E_2}$is along with BP.

From the figure, $\cos \theta =\frac{1}{\sqrt{5}}\text{and}\sin \theta =\frac{2}{\sqrt{5}}$

$$\begin{gathered} E_{2x}=E_2\cos \theta =0.432\times {10}^4\times \frac{1}{\sqrt{5}}=0.193\times {10}^4\mathrm{N}/\mathrm{C} \\ \mathrm{And}{\mathrm{E}}_{2\mathrm{y}}={\mathrm{E}}_2\mathrm{sin\theta }=0.432\times {10}^4\times \frac{2}{\sqrt{5}}=0.386\times {10}^4 \end{gathered}$$

So, the net electric field at P has both parallel and perpendicular components.

Therefore, the components of the net electric field at P is given by,

$$\begin{gathered} E_x=E_1+E_{2x}=(1.08+0.193)\times {10}^4=1.273\times {10}^4\mathrm{N}/\mathrm{C} \\ {\mathrm{E}}_{\mathrm{y}}={\mathrm{E}}_{2\mathrm{y}}=0.386\times {10}^4\mathrm{N}/\mathrm{C} \end{gathered}$$

So, the net electric field at P is given by,

$E={\sqrt{{{E^2}_x}^{}+{E^2}_y}}^{}=1.33\times {10}^{4}N/C$

The directive of the net electric field at P is given by,

Since

$$\begin{gathered} \cos \theta =\frac{E_x}{E} \\ \cos \theta =0.96 \\ \theta ={\cos }^{-1}\left(0.96\right) \\ =18° \end{gathered}$$