Question

Positive charge Q is distributed uniformly along the positive y-axis between y = 0 and y = a. A negative point charge -q lies on the positive x-axis, a distance x from the origin (in Fig.).

(a) Calculate the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis.

(b) Calculate the x- and y-components of the force that the charge distribution Q exerts on q.

(c) Show that if x W a, Fx _ -Qq>4pP0x2 and Fy _ +Qqa>8pP0x3 . Explain why this result is obtained.

Short answer

a) the x- and y-components of the electric field produced by the charge distribution Q at points on the positive x-axis is

$\overline{\mathrm{E}}=\frac{\mathrm{kQ}}{\mathrm{x}\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}}{\mathrm{i}}_{\mathrm{x}}-\frac{\mathrm{kQ}}{\mathrm{a}}\left[\frac{1}{\mathrm{x}}-\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2]{\mathrm{i}}_{\mathrm{y}}}}\right]{\mathrm{i}}_{\mathrm{y}}$

b) the x- and y-components of the force that the charge distribution Q exerts on q

$\overline{\mathrm{E}}=\frac{\mathrm{kQ}}{\mathrm{x}\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2}}{\mathrm{i}}_{\mathrm{x}}-\frac{\mathrm{kQ}}{\mathrm{a}}\left[\frac{1}{\mathrm{x}}-\frac{1}{\sqrt{{\mathrm{a}}^2+{\mathrm{x}}^2]{\mathrm{i}}_{\mathrm{y}}}}\right]{\mathrm{i}}_{\mathrm{y}}$

c) This is proven in equation (4) which is given below

Step-by-step solution

Step 1:

The illustration is shown in the figure below

We will donate the line density of the charge as

$\lambda =\frac{Q}{a}$

Integrating and substituting the given equation and its product to get a final desirable equation

a) Each dQ will generate dE which has two-component

$\mathrm{d}\overline{\mathrm{E}}=\frac{\mathrm{kdQ}}{{\mathrm{r}}^2}\cos (\mathrm{\theta }){\mathrm{i}}_{\mathrm{x}}$ (1)

a) Each dQ will generate dE which has two-component

$d{\overrightarrow{E}}_x=\frac{kdQ}{r^2}\cos \left(\theta \right)i_x$

Where: $\mathrm{dQ}=\mathrm{\lambda }-\mathrm{dy}$and $\cos (\theta )=x/\sqrt{x^2+y^2}$

Substitution in (1) to get

$\mathrm{d}\overline{{\mathrm{E}}_{\mathrm{z}}}=\frac{\mathrm{kx\lambda dy}}{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^{3/2}}{\mathrm{i}}_{\mathrm{z}}$ (2)

Integrate (7) to get that

$$\begin{gathered} {\overline{E}}_z={\int }_0^a\frac{\mathrm{kx\lambda dy}}{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^{3/2}}{\mathrm{i}}_{\mathrm{z}} \\ =\mathrm{k\lambda x}{{\left[\frac{\mathrm{y}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}\right]}_0}^{\mathrm{a}}\mathrm{ix} \\ =\frac{\mathrm{k\lambda x}}{\mathrm{x}\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}{\mathrm{i}}_{\mathrm{z}} \end{gathered}$$

Similarly, the vertical component is given by

$\mathrm{d}{\overline{\mathrm{E}}}_{\mathrm{x}}=\frac{\mathrm{kdQ}}{{\mathrm{r}}^2}\cos (\mathrm{\theta }){\mathrm{i}}_{\mathrm{x}}$

Where: $dQ=\lambda$and $\sin \left(\theta \right)=y/\sqrt{x^2+y^2}$

Substitution into the previous equation get

localid="1664530330650" $d{\overline{E}}_y=-\frac{ky\lambda dy}{{\left(x^2+y^2\right)}^{3/2}}{i}_y$

Integrate (2) to get that

$$\begin{gathered} {\overline{E}}_y={\int }_0^{\mathrm{a}}a-\frac{ky\lambda dy}{{\left(x^2+y^2\right)}^{3/2}}{i}_y \\ =-k\lambda {{\left[\frac{1}{x^2+y^2}\right]}_0}^a{i}_y \\ =-k\lambda \left[\frac{1}{x^2+a^2}-\frac{1}{x}\right]i_y \end{gathered}$$

The net electric field is given by

$$\begin{gathered} {\overline{E}}_y=\frac{k\lambda a}{x\sqrt[]{x^2+y^2}}{i}_x=-k\lambda {{\left[\frac{1}{x^2+a^2}+\frac{1}{x}\right]}_0}^a{i}_y \\ =\frac{kQ}{x\sqrt{x^2+a^2}}{i}_z-\frac{kQ}{a}\left[\frac{1}{x}-\frac{1}{x^2+a^2}\right]i_y \end{gathered}$$

b) the force is acting on the negative charge is in opposite direction to the electric field since the charge is negative

$=\frac{kQ}{x\sqrt{x^2+a^2}}{i}_z-\frac{kQ}{a}\left[\frac{1}{x}-\frac{1}{x^2+a^2}\right]i_y$

c) If the x? a then

$F_x=-\frac{kqQ}{x^2}=-\frac{qQ}{4\pi {\epsilon }_0{x}^2}$

Where : $\sqrt{x^2+a^2\approx x}$

Also the vertical component

$F_y=\frac{kqQ}{a}\left[\frac{\sqrt{x^2+a^2}-x}{x\sqrt{x^2+a^2}}\right]=\frac{kqQx}{a}\left[\frac{\sqrt{1+{\left(a/x\right)}^2}-1}{x\sqrt{x^2+a^2}}\right]$ (3)

Simplify using polynomial

$\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}=1+\frac{{\mathrm{a}}^2}{2{\mathrm{x}}^2}$

Substitution in (4) gives

$F_y=\frac{kqQ}{a}\left[\frac{a^2}{2x^2}\right]=\frac{qQx}{4{\mathrm{\pi \epsilon x}}^3}$ (4)

This result is expected because if x then we consider line charge Q as a point charge placed at x=0 and y=a/2 while calculating the force if we calculate the force using the approximate we will get the same result.