91影视

Draw the Cartesian coordinate system for the point $\left(x_0,y_0,z_0\right)$.

From the above figure, write the position vector.

$\mathit{S}\mathbf{=}\sqrt{{\mathbf{x}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{+}{\mathbf{y}}_{\mathbf{0}}^{\mathbf{2}}}$

Here,$x_0$and$y_0$are the two points on X and Y axis.

(a)

Write the expression for the electric field of charged conductor on z-axis.

$E_0=\frac{位}{2蟺{蔚}_{0}S}\hat{S}$ 鈥︹�� (1)

Here,$位$ is the line charge density and$\hat{S}$ is the unit vector.

Write the expression for the unit vector.

$\hat{S}=\frac{x_0\hat{x}+y_0\hat{y}}{\sqrt{x_0^2+y_0^2}}$

Substitute$\hat{S}=\frac{x_0\hat{x}+y_0\hat{y}}{\sqrt{x_0^2+y_0^2}}$ and$S=\sqrt{x_0^2+y_0^2}$ in equation (1).

role="math" localid="1654068317146" $$\begin{gathered} {\mathit{E}}_0=\frac{位}{2蟺{蔚}_0\left(\sqrt{x_0^2+y_0^2}\right)}\left(\frac{x_0\hat{x}+y_0\hat{y}}{\sqrt{x_0^2+y_0^2}}\right) \\ {\mathit{E}}_0=\frac{位}{2蟺{蔚}_0\left(x_0^2+y_0^2\right)}\left(x_0\hat{x}+y_0\hat{y}\right) \end{gathered}$$ 鈥︹�� (2)

Therefore, the electric field in Cartesian coordinates for the point$\left(x_0,y_0,z_0\right)$ is role="math" localid="1654068403951" ${\mathit{E}}_0=\frac{位}{2蟺{蔚}_0\left(x_0^2+y_0^2\right)}\left(x_0\hat{x}+y_0\hat{y}\right)$.

(b)

Write the expressions for an inverse Lorentz transform equations.

$$\begin{gathered} x_0=纬\left(x+vt\right) \\ {y}_0=y \\ {\stackrel{鈬赌}{E}}_x=E_x \\ {\stackrel{鈬赌}{E}}_y=纬E_y \end{gathered}$$

Here,$纬$ is the Lorentz contraction, v is the speed and t is the time.

Write equation (2) in the form of x component of an electric field.

$E_x=\frac{位}{2蟺{蔚}_0\left(x_0^2+y_0^2\right)}{x}_0$

As it is known that ${\stackrel{鈬赌}{E}}_x=E_x$, the value of${\stackrel{鈬赌}{E}}_x$will be,

${\stackrel{鈬赌}{E}}_x=\frac{位}{2蟺{蔚}_0\left(x_0^2+y_0^2\right)}{x}_0$

Substitute$x_0=纬\left(x+vt\right)$and$y_0=y$in the above expression.

${\stackrel{鈬赌}{E}}_x=\frac{位}{2蟺{蔚}_0\left({\left(纬\left(x+vt\right)\right)}^2+y^2\right)}纬\left(x+vt\right)$

Similarly, write equation (2) in the form of y-component of an electric field.

$E_y=\frac{位}{2蟺{蔚}_0\left(x_0^2+y_0^2\right)}{y}_0$

Also, role="math" localid="1654068933653" ${\stackrel{鈬赌}{E}}_y=纬E_y$the value of${\stackrel{鈬赌}{E}}_y$will be,

$\stackrel{鈬赌}{E_y}=纬\frac{位}{2蟺{蔚}_0\left(x_0^2+y_0^2\right)}{y}_0$

Substitute$x_0=纬\left(x+vt\right)$and$y_0=y$in the above expression.

role="math" localid="1654069100586" ${\stackrel{鈬赌}{E}}_y=纬\frac{位}{2蟺{蔚}_0\left({\left(纬\left(x+vt\right)\right)}^2+y^2\right)}y$

Hence, the net electric field will be,

$$\begin{gathered} \stackrel{鈬赌}{E}={\stackrel{鈬赌}{E}}_x+{\stackrel{鈬赌}{E}}_y \\ \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0\left({\left(纬\left(x+vt\right)\right)}^2+y^2\right)}纬\left(x+vt\right)+纬\frac{纬}{2蟺{蔚}_0\left({\left(纬\left(x+vt\right)\right)}^2+y^2\right)}y \\ \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0\left({\left(纬\left(x+vt\right)\right)}^2+y^2\right)}\left(纬\left(x+vt\right)+纬y\right) \end{gathered}$$

On further solving,

$$\begin{gathered} \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0}\left(\frac{纬\left(x+vt\right)+纬y}{纬{\left(x+vt\right)}^2+y^2}\right) \\ \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0}\frac{1}{纬}\left(\frac{\left(x+vt\right)\hat{x}+y\hat{y}}{{\left(x+vt\right)}^2+{\left({\displaystyle \frac{y}{纬}}\right)}^2}\right) \end{gathered}$$ 鈥︹�� (3)

Here,$\left(x+vt\right)\hat{x}+y\hat{y}=S$

Take the magnitude of S.

$$\begin{gathered} S=\sqrt{{\left(x+vt\right)}^2+y^2} \\ {S}^2={\left(x+vt\right)}^2+y^2 \end{gathered}$$

Here, $y=S\sin 胃$.

Solve the denominator value of equation (3).

$$\begin{gathered} {\left(x+vt\right)}^2+{\left(\frac{y}{纬}\right)}^2={\left(x+vt\right)}^2+y^2\left(1-\frac{v^2}{c^2}\right) \\ {\left(x+vt\right)}^2+{\left(\frac{y}{纬}\right)}^2={\left(x+vt\right)}^2+y^2-\frac{y^2{v}^2}{c^2} \\ {\left(x+vt\right)}^2+{\left(\frac{y}{纬}\right)}^2=S^2-{\left(S\sin 胃\right)}^2\frac{v^2}{c^2} \\ {\left(x+vt\right)}^2+{\left(\frac{y}{纬}\right)}^2=S^2-{\left(\frac{v}{c}\right)}^2{S}^2{\sin }^2胃 \end{gathered}$$

Substitute $\frac{1}{{纬}^2}=1-\frac{v^2}{c^2},\left(x+vt\right)\hat{x}+y\hat{y}=S$and ${\left(x+vt\right)}^2+{\left(\frac{y}{纬}\right)}^2=S^2-{\left(\frac{v}{c}\right)}^2{S}^2{\sin }^2胃$in equation (3).

$$\begin{gathered} \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0}\left(\sqrt{1-\frac{v^2}{c^2}}\right)\left(\frac{S}{S^2-{\left({\displaystyle \frac{v}{c}}\right)}^2{S}^2{\sin }^2胃}\right) \\ \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0}\left(\sqrt{1-\frac{v^2}{c^2}}\right)\left(\frac{S}{S^2\left(1-{\left({\displaystyle \frac{v}{c}}\right)}^2{\sin }^2胃\right)}\right) \\ \stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0}\left(\frac{\left(\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}\right)}{S\left(1-{\left({\displaystyle \frac{v}{c}}\right)}^2{\sin }^2胃\right)}\right) \end{gathered}$$

Hence, the field point is away from the instantaneous location of the wire.

Therefore, the electric field in S is $\stackrel{鈬赌}{E}=\frac{位}{2蟺{蔚}_0}\left(\frac{\left(\sqrt{1-{\displaystyle \frac{v^2}{c^2}}}\right)}{S\left(1-{\left({\displaystyle \frac{v}{c}}\right)}^2{\sin }^2胃\right)}\right)$.