Figure 22-37 shows two charged particles on an x-axis: $\mathbf{-}\mathbf{q}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}\mathbf{C}$at $\mathbf{x}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$$\mathbf{q}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{20}\mathbf{\times }{\mathbf{10}}^{\mathbf{-}\mathbf{19}}$and at $\mathit{x}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{.}\mathbf{00}\mathbf{}\mathbf{m}$. What are the (a) magnitude and (b) direction (relative to the positive direction of the x-axis) of the net electric field produced at point Pat $\mathbf{y}\mathbf{=}\mathbf{4}\mathbf{.}\mathbf{00}\mathbf{m}$?

1. The two charges on the x-axis are $-\mathrm{q}=-3.20\times {10}^{-19}\mathrm{C}$and $\mathrm{q}=3.20\times {10}^{-19}$placed at $-x=-3.00\mathrm{m}$and $x=3.00\mathrm{m}$.
2. The point is placed on the y-axis at y= 4 m

Using the concept of the electric field at a given point, we can get the value of an individual electric field by a charge. Again using the superposition law, we can get the value of the electric field in its direction and this determines the net electric field at that point.

Formulae:

The magnitude of the electric field,$\mathrm{E}=\frac{q}{4{\mathrm{\pi \epsilon }}_0{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^{3/2}}$ (1)

where R = The distance of field point from the charge, and q = charge of the particle

According to the superposition principle, the electric field at a point due to more than one charge,

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\sum _{\mathrm{i}=1}^{\mathrm{n}}\overrightarrow{\mathrm{E}} \\ =\sum _{\mathrm{i}=1}^{\mathrm{n}}\frac{{\mathrm{q}}_{\mathrm{i}}}{4{\mathrm{\pi \epsilon }}_0{{\mathrm{r}}_{\mathrm{i}}}^2}{\hat{\mathrm{r}}}_{\mathrm{i}} \end{gathered}$$ (2)

The vertical components of the two charges at the point cancel each other due to symmetry (opposite directions), while using x=300m and y=4.00 m, the horizontal component both pointing towards negative x-axis adds up considering equation (2) and thus, gives a net electric field using equation (1) as follows:

$$\begin{gathered} \overrightarrow{\mathrm{E}}=\frac{2\left|\mathrm{q}\right|\mathrm{d}}{4\mathrm{\pi \epsilon }{\left({\mathrm{d}}^2+{\mathrm{y}}^2\right)}^{3/2}} \\ =\frac{2\left(8.99\times {10}^9\mathrm{N}.{\displaystyle \frac{{\mathrm{m}}^2}{{\mathrm{c}}^2}}\right)\left(3.20\times {10}^{-19}\mathrm{C}\right)\left(3.00\mathrm{m}\right)}{{\left[{\left(3.00\mathrm{m}\right)}^2+{\left(4.00\mathrm{m}\right)}^2\right]}^{{\displaystyle \frac{3}{2}}}} \\ =1.38\times {10}^{-10}\frac{\mathrm{N}}{\mathrm{C}} \end{gathered}$$

Hence, the value of the net electric field is$1.38\times {10}^{-10}\frac{\mathrm{N}}{\mathrm{C}}$

From part (a) calculations, the direction of the field is seen to be pointing towards the negative x-axis or is in $180°$counter-clockwise direction of the +x-axis.