Q37P - Introduction to Electrodynamics

Question

Find the interaction energy $\left({\in }_0\int E_1.E_2 d\tau {\in }_0\int E_1-E_{2}d\tau in Eq.2.47\right)$

for two point

charges $q_1$and $q_2$a distance a apart.

Step-by-Step Solution

Answer

The interaction energy is $W=\frac{q_1{q}_2}{4{\mathrm{\pi \epsilon }}_0\mathrm{a}}.$

Step 1: Determine the expression for electric fields.

Write the expression for the interaction energy.

$W={\epsilon }_0\int E_1-E_{2}d\tau$

Here, $E_1$is the electric field due to charge$q_{1 . }{E}_2$is the electric field due to charge$q_{2 }. d\tau$

is the volume element, and ${\epsilon }_0$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, K is the Coulomb’s constant, r is 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 charge$q_2$.

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

Here, R is the distance from the charge, and$\hat{R}$is the unit vector that represents the direction of an electric field due to charge$q_2$.

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

                           

Here,$\theta$is the angle made by the electric$E_1$field with the z axis and$\beta$is the angle between the two electric field vectors at point P.

Step 2: Determine value of R

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

$R=\sqrt{r^2+a^2-2ar c\mathrm{os}\theta }$

Squaring on the side 

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

Step 3: Determine the value of the interaction energy.

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:

$\cos \beta =\frac{r-a\cos \theta }{R}$

Rewrite the interaction energy term in the following way:

$W={\epsilon }_0\int E_1{E}_2 \cos \theta \beta d\tau$

Substitute $\frac{k{q}_1}{r^2}$for $E_1, \frac{k{q}_2}{R^2}$for $E_2$and $r^{2}dr \sin \theta d\theta d\varnothing$for $d\tau$and $\frac{r-a \cos \theta }{R}$for $\cos \beta$ in the equation.

$$\begin{gathered} W={\epsilon }_0\int \left(\begin{array}{c}\frac{k{q}_1}{r^2}\end{array}\right)\left(\begin{array}{c}\frac{k{q}_2}{r^2}\end{array}\right)\left(\begin{array}{c}\frac{r-\cos \theta }{R}\end{array}\right)r^{2}dr \sin \theta d\theta d\varphi \\ ={\epsilon }_0{K}^2{q}_1{q}_2\int \left(\begin{array}{c}\frac{r-\cos \theta }{R}\end{array}\right)dr \sin \theta d\theta d\varphi \end{gathered}$$

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

$$\begin{gathered} W={\epsilon }_0{K}^2{q}_1{q}_2\int \left(\begin{array}{c}\frac{r-\cos \theta }{R}\end{array}\right)r^{2}dr \sin \theta d\theta d\varphi \\ ={\epsilon }_0{K}^2{q}_1{q}_2\int \int \int \left(\begin{array}{c}\frac{dR}{R^2}\end{array}\right) (\sin \theta d\theta d\varphi ) \end{gathered}$$

Integrate the above equation,

$$\begin{gathered} W={\epsilon }_0{K}^2{q}_1{q}_2\int \left(\begin{array}{c}\frac{aR}{R^2}\end{array}\right){\int }_0 \sin \theta d\theta {\int }_{0}d\varphi \\ ={\epsilon }_0{K}^2{q}_1{q}_2\left(\begin{array}{c}\frac{1}{\infty }-\left(\begin{array}{c}-\frac{1}{a}\end{array}\right)\end{array}\right)(-\mathrm{cos\pi }-(-\cos 0\left)\right) (2\mathrm{\pi }-0) \\ = \frac{{\mathrm{\epsilon }}_0{\mathrm{K}}^2{\mathrm{q}}_1{\mathrm{q}}_2}{\mathrm{a}}\left(2\right)\left(2\mathrm{\pi }\right) \end{gathered}$$

Substitute $\frac{1}{4{\mathrm{\pi \epsilon }}_0}$ for K in above equation.

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

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