A parallel-plate capacitor, at rest in ${\mathbf{S}}_{\mathbf{0}}$and tilted at a $\mathbf{45}\mathbf{°}$angle to the ${\mathit{x}}_{\mathbf{0}}$axis, carries charge densities $\mathbf{\pm }{\mathit{\sigma }}_{\mathbf{0}}$on the two plates (Fig. 12.41). System$\mathit{S}$ is moving to the right at speed $\mathit{V}$ relative to ${\mathit{S}}_{\mathbf{0}}$.

(a) Find ${\mathit{E}}_{\mathbf{0}}$, the field in ${\mathit{S}}_{\mathbf{0}}$.

(b) Find $\mathit{E}$, the field in $\mathit{S}$.

(c) What angle do the plates make with the $\mathit{x}$axis?

(d) Is the field perpendicular to the plates in $\mathit{S}$?

The parallel plates of capacitor are tiled at angle is${\theta }_0=45°$ to the $X_0$axis.

Charge density of the capacitor plates are $\pm {\sigma }_0$.

The system $S$is moving right relative to$\overline{S}$ at speed of$v$ .

The expression to calculate the magnitude of the electric field is given as follows.

${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{\sigma }}_{\mathbf{0}}}{{\mathbf{\epsilon }}_{\mathbf{0}}}$ …… (1)

Here, ${\mathbf{\epsilon }}_{\mathbf{0}}$is the permittivity of space.

The expression to calculate the angle made by the parallel plates with$x$ axis is given as follows.

$\mathit{t}\mathit{a}\mathit{n}\mathit{\theta }\mathbf{=}\frac{\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{\theta }}_{\mathbf{0}}}{\frac{\mathbf{1}}{\mathbf{\gamma }}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{\theta }}_{\mathbf{0}}}$ …… (2)

The expression calculates the angle between the normal plates and electric field is given as follows.

$\mathbf{cos\varphi }\mathbf{=}\frac{\mathbf{E}\mathbf{\times }\widehat{\mathbf{n}}}{\mathbf{\left|}\mathbf{E}\mathbf{\right|}}$ …… (3)

(a)

The electric field in vector form in frame $S_0$is given by,

${\text{E}}_{\text{0}}=-E_0\cos 45°\widehat{x}+E_0\sin 45°\widehat{y}$

Substitute $\frac{{\sigma }_0}{{\epsilon }_0}$for $E_0$into above equation.

$\begin{array}{l}{\text{E}}_0=-\frac{{\sigma }_0}{{\epsilon }_0}\cos 45°\widehat{x}+\frac{{\sigma }_0}{{\epsilon }_0}\sin 45°\widehat{y}\\ {\text{E}}_0=-\frac{{\sigma }_0}{{\epsilon }_0}\frac{1}{\sqrt{2}}+\frac{{\sigma }_0}{{\epsilon }_0}\frac{1}{\sqrt{2}}\widehat{y}\\ {\text{E}}_0=\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}(-\widehat{x}+\gamma \widehat{y})\end{array}$

Hence the electrical field$E_0$ in frame$S_0$ is$\frac{{\mathrm{\sigma }}_0}{\sqrt{2}{\mathrm{\epsilon }}_0}(-\widehat{\mathrm{x}}+\mathrm{\gamma }\widehat{\mathrm{y}})$ .

(b)

The $x$component of the electric field is given by,

$\begin{array}{l}{E}_x=E_{x0}\\ E_x=-\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}\widehat{x}\end{array}$

The$y$component of the electric field is given by,

role="math" localid="1658297161533" $\begin{array}{l}{E}_y=\gamma E_y\\ E_y=\gamma \frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}\widehat{y}\end{array}$

The electric field $E$in the frame $S$ is given by,

$E=E_x+E_y$

Substitute$-\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}$ for $E_x$and$\gamma \frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}$ for$E_y$ into above equation.

$\begin{array}{l}E=-\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}\widehat{x}+\gamma \frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}\widehat{y}\\ E=\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}(-\widehat{x}+\gamma \widehat{y})\end{array}$

Hence the electric field$E$ in the frame$S$ is$\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}(-\widehat{x}+\gamma \widehat{y})$ .

(c)

Calculate the angle made by plates with $x$axis.

Substitute$45°$ for$\theta$ into equation (2).

$\begin{array}{c}\tan \theta =\frac{\sin 45°}{\frac{1}{\gamma }\cos 45°}\\ \tan \theta =\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\gamma }\frac{1}{\sqrt{2}}}\\ \tan \theta =\gamma \\ \theta ={\tan }^{-1}\left(\gamma \right)\end{array}$

Hence the angle made by plates with$x$ axis is${\tan }^{-1}\left(\gamma \right)$ .

(d)

Let assume $\widehat{n}$ is the vector which is perpendicular to the frame $S$.

The vector$\widehat{n}$is given by,

$\widehat{n}=-\sin \theta \widehat{x}+\cos \theta \widehat{y}$

Calculate the angle between normal to plates and electric field.

Substitute $\left(\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}(-\widehat{x}+\gamma \widehat{y})\right)$for $E$, $-\sin \theta \widehat{x}+\cos \theta \widehat{y}$ for $\hat{n}$ and $\frac{{\sigma }_0}{\sqrt{2}{\epsilon }_0}\sqrt{1+{\gamma }^2}$for $\left|E\right|$into equation (3).

$\begin{array}{l}\mathrm{cos\varphi }=\frac{\left(\frac{{\mathrm{\sigma }}_0}{\sqrt{2}{\mathrm{\epsilon }}_0}(-\widehat{\mathrm{x}}+\mathrm{\gamma }\widehat{\mathrm{y}})\right)\cdot \left(-\mathrm{sin\theta }\widehat{\mathrm{x}}+\mathrm{cos\theta }\widehat{\mathrm{y}}\right)}{\frac{{\mathrm{\sigma }}_0}{\sqrt{2}{\mathrm{\epsilon }}_0}\sqrt{1+{\mathrm{\gamma }}^2}}\\ \mathrm{cos\varphi }=\frac{(-\widehat{\mathrm{x}}+\mathrm{\gamma }\widehat{\mathrm{y}})\cdot (-\mathrm{sin\theta }\widehat{\mathrm{x}}+\mathrm{cos\theta }\widehat{\mathrm{y}})}{\sqrt{1+{\mathrm{\gamma }}^2}}\\ \mathrm{cos\varphi }=\frac{\mathrm{sin\theta }+\mathrm{\gamma cos\theta }}{\sqrt{1+{\mathrm{\gamma }}^2}}\\ \mathrm{cos\varphi }=\frac{\mathrm{cos\theta }(\mathrm{tan\theta }+\mathrm{\gamma })}{\sqrt{1+{\mathrm{\gamma }}^2}}\end{array}$ …… (4)

Resolve as:

$\begin{array}{c}\tan \theta =\gamma \\ \frac{\sin \theta }{\cos \theta }=\gamma \\ \frac{\sqrt{{\cos }^2\theta -1}}{\cos \theta }=\gamma \\ \sqrt{\frac{1}{{\cos }^2\theta }-1}=\gamma \end{array}$

Solve further as,

$\begin{array}{c}\frac{1}{{\cos }^2\theta }-1={\gamma }^2\\ \frac{1}{{\cos }^2\theta }={\gamma }^2+1\\ {\cos }^2\theta =\frac{1}{{\gamma }^2+1}\\ \cos \theta =\frac{1}{\sqrt{{\gamma }^2+1}}\end{array}$

Substitute $\frac{1}{\sqrt{{\gamma }^2+1}}$ for $cos\theta$and $\gamma$for $\tan \theta$into equation (4).

$\begin{array}{l}\cos \varphi =\frac{1}{\sqrt{1+{\gamma }^2}}\frac{(\gamma +\gamma )}{\sqrt{1+{\gamma }^2}}\\ \cos \varphi =\frac{2\gamma }{1+{\gamma }^2}\end{array}$

The angle between normal to plates and field is not zero. Therefore, the electric field is not perpendicular to the plates.