Question

In Fig. 22-36, the four particles are fixed in place and have charges,${\mathbf{q}}_{\mathbf{1}}\mathbf{=}{\mathbf{q}}_{\mathbf{2}}\mathbf{=}\mathbf{+}\mathbf{5}\mathbf{e}\mathbf{}\mathbf{,}{\mathbf{q}}_{\mathbf{3}}\mathbf{}\mathbf{=}\mathbf{+}\mathbf{3}\mathbf{e}\mathbf{}\mathbf{and}\mathbf{}{\mathbf{q}}_{\mathbf{4}}\mathbf{=}\mathbf{-}\mathbf{12}\mathbf{e}$. Distance, d = 5.0 mm. What is the magnitude of the net electric field at point Pdue to the particles?

Short answer

The magnitude of the net electric field at point P due to the particle is 0.

Step-by-step solution

The given data

1. The charges of the four particles,${\mathrm{q}}_1={\mathrm{q}}_2=+5\mathrm{e},{\mathrm{q}}_3=+3\mathrm{e}\mathrm{and}{\mathrm{q}}_4=-12\mathrm{e}$
2. The distance,$\mathrm{d}=5\mathrm{mm}\left(\frac{1\mathrm{m}}{1000\mathrm{mm}}\right)=0.005\mathrm{m}$

Understanding the concept of electric field 

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,$\mathrm{E}=\frac{\mathrm{q}}{4{\mathrm{\pi \epsilon }}_0{\mathrm{R}}^2}\hat{\mathrm{R}}$ (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,

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

Calculation of the net electric field

The origin of the coordinate system is placed at point P and the y-axis is oriented in the direction of the charge, ${\mathrm{q}}_4=-12\mathrm{e}$(passing through the charge, ${\mathrm{q}}_3=+3\mathrm{e}$). The x-axis is perpendicular to the y axis, and thus passes through the identical charges,

${\mathrm{q}}_1={\mathrm{q}}_2=+5\mathrm{e}$

The individual magnitudes of the electric field due to the charges are figured by using the absolute signs of the charges. Now, considering the point charge being positive ( q > 0), we can see that the contributions coming from them cancel each other. Hence, the net electric field in the direction of the y-axis is given using equations (1) and (2) as follows:

$$\begin{gathered} \overrightarrow{E_{net}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{\left|q_4\right|}{{\left(2d\right)}^2}-\frac{q_3}{{\left(d\right)}^2}\right]\hat{j} \\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\left[\frac{12q}{4d^2}-\frac{3q}{d^2}\right]\hat{j} \\ =0 \end{gathered}$$

Hence, the value of the net electric field is 0. The rough sketch of the field lines is given below: