91影视

The illustration is shown in the figure below

We will donate the line density of the charge as

$位=\frac{Q}{a}$

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

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

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

$d{\stackrel{鈫�}{E}}_x=\frac{kdQ}{r^2}\cos 鈦�\left(胃\right)i_x$

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

Substitution in (1) to get

$\mathrm{d}\stackrel{炉}{{\mathrm{E}}_{\mathrm{z}}}=\frac{\mathrm{办虫位诲测}}{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^{3/2}}{\mathrm{i}}_{\mathrm{z}}$ (2)

Integrate (7) to get that

$$\begin{gathered} {\stackrel{炉}{E}}_z={鈭�}_0^a\frac{\mathrm{办虫位诲测}}{{\left({\mathrm{x}}^2+{\mathrm{y}}^2\right)}^{3/2}}{\mathrm{i}}_{\mathrm{z}} \\ =\mathrm{办位虫}{{\left[\frac{\mathrm{y}}{{\mathrm{x}}^2\sqrt{{\mathrm{x}}^2+{\mathrm{y}}^2}}\right]}_0}^{\mathrm{a}}\mathrm{ix} \\ =\frac{\mathrm{办位虫}}{\mathrm{x}\sqrt{{\mathrm{x}}^2+{\mathrm{a}}^2}}{\mathrm{i}}_{\mathrm{z}} \end{gathered}$$

Similarly, the vertical component is given by

$\mathrm{d}{\stackrel{炉}{\mathrm{E}}}_{\mathrm{x}}=\frac{\mathrm{kdQ}}{{\mathrm{r}}^2}\cos (\mathrm{胃}){\mathrm{i}}_{\mathrm{x}}$

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

Substitution into the previous equation get

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

Integrate (2) to get that

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

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

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

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

$F_x=-\frac{kqQ}{x^2}=-\frac{qQ}{4蟺{蔚}_0{x}^2}$

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

$F_y=\frac{kqQ}{a}\left[\frac{a^2}{2x^2}\right]=\frac{qQx}{4{\mathrm{蟺蔚虫}}^3}$ (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.