A particle of charge$\mathbf{-}{\mathit{q}}_{\mathbf{1}}$is at the origin of an xaxis. (a) At what location on the axis should a particle of charge$\mathbf{-}\mathbf{4}{\mathit{q}}_{\mathbf{1}}$be placed so that the net electric field is zero at$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{2}\mathbf{.0}\mathbf{}\mathit{m}\mathit{m}$on the axis? (b) If, instead, a particle of charge$\mathbf{+}\mathbf{4}{\mathit{q}}_{\mathbf{1}}$is placed at that location, what is the direction (relative to the positive direction of the xaxis) of the net electric field at$\mathit{x}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{}\mathit{m}\mathit{m}$?

1. A particle of charge$-q_1$is at the origin of an x axis.
2. A particle of charge$-4q_1$is placed so that the net electric field is zero at$x=2.0mm$on the axis.
3. A particle of charge$+4q_1$ is placed so that the net electric field is zero at $x=2.0mm$on the axis

Using the concept of the electric field due to a particle at a given point, we can get the values of the individual electric fields of the charges. Thus, in the case of the net-zero fields, the two fields balance each other and cancel. Similarly, using the same concept the direction of the net electric field can be given.

Formula:

The electric field due to a charge particle at a given point, $\overrightarrow{E}=\frac{q}{4\pi {\epsilon }_o{r}^2}\widehat{r}$ (iv)

Where,

r = the distance of field point from the charge

q = charge of the particle

Since the two charges in question are of the same sign, the point should be located in between them (so that the field vectors point in the opposite direction). Let the coordinate of the second particle be$x\text{'}(x\text{'}>0)$

Then, the magnitude of the field due to the charge$-q_1$evaluated at$x$is given by using equation (i) as:

$E=\text{-}{q}_1/4\pi {\epsilon }_o{x}^2\mathrm{..................}\left(a\right)$

While, the electric field due to the second charge -4q₁ is given using equation (i) as:

$E\text{'}=4q_1/4\pi {\epsilon }_o{(x\text{'}–x)}^2\mathrm{............}\left(b\right)$

To get the net electric field as zero, we equate both the equations (a) and (b) to get the location of the particle on the x-axis as follows:

$\begin{array}{c}\left|E\right|=|E\text{'}|\\ \frac{q_1}{4\pi {ϵ}_o{x}^2}=\frac{4q_1}{4\pi {ϵ}_o{|x^{\text{'}}-x|}^2}\\ \frac{|x^{\text{'}}-x|}{x}=2\\ x^{\text{'}}=3x\\ =3\left(2mm\right)\\ =6mm\end{array}$

Thus, the value of the location of the particle is.$6mm$

In this case, with the second charge now positive, the electric field vectors produced by both charges are in the negative x direction, when evaluated at x = 2.0 mm. Therefore, the net field points in the negative x direction, or , measured counter-clockwise from the +x axis.