91影视

Write the expression for the interaction energy.

$W={蔚}_0鈭�E_1-E_{2}d蟿$

Here, $E_1$is the electric field due to charge$q_{1.}{E}_2$is the electric field due to chargelocalid="1654680864961" $q_2.d蟿$

is the volume element, and ${蔚}_0$is the permittivity of free space.

Write the expression for the electric field due to charge.

${\mathit{E}}_{\mathbf{1}}\mathbf{=}\frac{\mathbf{K}{\mathbf{q}}_{\mathbf{1}}}{{\mathbf{r}}^{\mathbf{2}}}\mathbf{(}\mathbf{}\hat{\mathbf{r}}\mathbf{}\mathbf{)}$

Here, Kis the Coulomb鈥檚 constant, ris the distance from the charge, and

$\hat{r}$is the unit vector that represents the direction of an electric field due to charge.

Write the expression for the electric field due to chargelocalid="1654685113455" $q_2$.

localid="1654685109004" ${\mathit{E}}_{\mathbf{2}}\mathbf{=}\frac{\mathbf{K}{\mathbf{q}}_{\mathbf{2}}}{{\mathbf{R}}^{\mathbf{2}}}\mathbf{(}\mathbf{}\hat{\mathbf{R}}\mathbf{}\mathbf{)}$

Here,Ris the distance from the charge, andlocalid="1654685092952" $\hat{R}$is the unit vector that represents the direction of an electric field due to chargelocalid="1654685088381" $q_2$.

Consider the diagram for the configuration is shown in figure below.

Here,$胃$is the angle made by the electric$E_1$field with the zaxis and$尾$is the angle between the two electric field vectors at point P.

The value of R in the figure above, according to the law of cosines, is,

$R=\sqrt{r^2+a^2-2arc\mathrm{os}胃}$

Squaring on the side

$$\begin{gathered} 2RdR=2rdr-2adr\cos 胃 \\ RdR=rdr-adr\cos 胃 \\ RdR=(r-a\cos 胃)dr \end{gathered}$$

Now to redraw the above figure,

Draw a perpendicular from the charge $q_2$as shown in above figure.

From above figure, the angle between the two electric field vectors' cosine is represented as follows:

$\cos 尾=\frac{r-a\cos 胃}{R}$

Rewrite the interaction energy term in the following way:

$W={蔚}_0鈭�E_1{E}_2\cos 胃尾d蟿$

Substitute $\frac{k{q}_1}{r^2}$for $E_1,\frac{k{q}_2}{R^2}$for $E_2$and $r^{2}dr\sin 胃d胃d鈭�$for $d蟿$and $\frac{r-a\cos 胃}{R}$for $\cos 尾$in the equation.

localid="1654684150710" $$\begin{gathered} W={蔚}_0鈭�\left(\begin{array}{c}\frac{k{q}_1}{r^2}\end{array}\right)\left(\begin{array}{c}\frac{k{q}_2}{r^2}\end{array}\right)\left(\begin{array}{c}\frac{r-\cos 胃}{R}\end{array}\right)r^{2}dr\sin 胃d胃d蠒 \\ ={蔚}_0{K}^2{q}_1{q}_2鈭�\left(\begin{array}{c}\frac{r-\cos 胃}{R}\end{array}\right)dr\sin 胃d胃d蠒 \end{gathered}$$

Now substitute RdR for $(r-a\cos 胃)dr$in the equation.

localid="1654684297608" $$\begin{gathered} W={蔚}_0{K}^2{q}_1{q}_2鈭�\left(\begin{array}{c}\frac{r-\cos 胃}{R}\end{array}\right)r^{2}dr\sin 胃d胃d蠒 \\ ={蔚}_0{K}^2{q}_1{q}_2鈭�鈭�鈭�\left(\begin{array}{c}\frac{dR}{R^2}\end{array}\right)(\sin 胃d胃d蠒) \end{gathered}$$

Integrate the above equation,

$$\begin{gathered} W={蔚}_0{K}^2{q}_1{q}_2鈭�\left(\begin{array}{c}\frac{aR}{R^2}\end{array}\right){鈭�}_0\sin 胃d胃{鈭�}_{0}d蠒 \\ ={蔚}_0{K}^2{q}_1{q}_2\left(\begin{array}{c}\frac{1}{鈭�}-\left(\begin{array}{c}-\frac{1}{a}\end{array}\right)\end{array}\right)(-\mathrm{肠辞蝉蟺}-(-\cos 0\left)\right)(2\mathrm{蟺}-0) \\ =\frac{{\mathrm{蔚}}_0{\mathrm{K}}^2{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{a}}\left(2\right)\left(2\mathrm{蟺}\right) \end{gathered}$$

Substitute $\frac{1}{4{\mathrm{蟺蔚}}_0}$ for Kin above equation.

$$\begin{gathered} W=\frac{{{蔚}_{0\left(\begin{array}{c}{\displaystyle \frac{1}{4{\mathrm{蟺蔚}}_0}}\end{array}\right)}}^2{q}_1{q}_2}{a}\left(2\right)\left(2\mathrm{蟺}\right) \\ =\frac{{\mathrm{q}}_1{\mathrm{q}}_2}{4{\mathrm{蟺蔚}}_0\mathrm{a}} \end{gathered}$$

Hence, the interaction energy is$W=\frac{q_1{q}_2}{4{\mathrm{蟺蔚}}_0\mathrm{a}}.$