Two particles, each of positive charge q, are fixed in place on a yaxis, one at$\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathit{d}$and the other at.$\mathit{y}\mathbf{}\mathbf{=}\mathbf{}\mathbf{-}\mathit{d}$ (a) Write an expression that gives the magnitude Eof the net electric field at points on the xaxis given by.$\mathit{x}\mathbf{}\mathbf{=}\mathbf{}\mathit{a}\mathit{d}$ (b) Graph Eversus$\mathit{\alpha }$for the range$\mathbf{0}\mathbf{<}\mathit{\alpha }\mathbf{<}\mathbf{}\mathbf{4}$. From the graph, determine the values of$\mathit{\alpha }$that give (c) the maximum value of Eand (d) half the maximum value of E.

Two particles, each of positive charge, are fixed in place on a y axis, one at $y=d$and the other at$y=-d$.

Using the concept of the electric field, we can get the required expression of the field. The value of the distance t which we get the maximum field is calculated by equating the differentiation of the electric field expression to zero. Similarly, the value of the distance is calculated for half the maximum electric field.

Formula:

The magnitude of the electric field, $E=\frac{q}{4\pi {\epsilon }_o{r}^2}\widehat{r}\text{where,}\widehat{r}=\cos \theta \widehat{i}+\sin \theta \widehat{j}$ (i)

where, r = The distance of field point from the charge

q = charge of the particle

Let$q_1$denote the charge at$y=d$and$q_2$denote the charge at.$y=-d$ The individual magnitudes$E_1$and$E_2$are figured from equation (i), where the absolute value signs for$q$are unnecessary since these charges are both positive. The distance from$q_1$to a point on the x-axis is the same as the distance fromto a point on the x-axis, which is given as:

$r=\sqrt{(x^2+d^2)}$

By symmetry, the y-component of the net field along the x axis is zero.

Thus, the x component of the net field, evaluated at points on the positive x-axis, is given as:

$E_x=2\left(\frac{1}{4\pi {ϵ}_o}\right)\left(\frac{q}{x^2+d^2}\right)\left(\frac{x}{{\sqrt{x^2+d^2}}^{}}\right)$

Where the last factor is$cos\theta =x/r$with$\theta$being the angle for each individual field as measured from the x axis.

Solving the above expression, by substituting the value,$x=\alpha d$ we obtain the electric field as:

$E_x=\frac{q}{4\pi {ϵ}_o{d}^2}\left(\frac{\alpha }{{({\alpha }^2+1)}^{3/2}}\right)$

Hence, the value of the expression of the electric field is.$\frac{q}{4\pi {ϵ}_o{d}^2}\left(\frac{\alpha }{{({\alpha }^2+1)}^{3/2}}\right)$

The graph of$E=E_x$versus$\alpha$is shown below. For the purposes of graphing, we set $d=1m$and.$q=5.56x{10}^{–11}C$

Hence, the required graph is plotted.

From the graph, we estimate the electric field$E_{max}$occurs at$\alpha =0.71$about.

Thus, the value at which we get the maximum field occur is1/√ 2.

The graph suggests that “half-height” points occur at$\alpha \approx 0.2$and$\alpha \approx 2.0$ .

Further numerical exploration the desired value of the electric field leads to the values of$\alpha$ as 0.2047 and 1.9864.