91影视

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

Charge density of the capacitor plates are $卤{蟽}_0$.

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

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

${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{蟽}}_{\mathbf{0}}}{{\mathbf{蔚}}_{\mathbf{0}}}$ 鈥︹�� (1)

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

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

$\mathit{t}\mathit{a}\mathit{n}\mathit{胃}\mathbf{=}\frac{\mathbf{c}\mathbf{o}\mathbf{s}{\mathbf{胃}}_{\mathbf{0}}}{\frac{\mathbf{1}}{\mathbf{纬}}\mathbf{s}\mathbf{i}\mathbf{n}{\mathbf{胃}}_{\mathbf{0}}}$ 鈥︹�� (2)

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

$\mathbf{肠辞蝉蠒}\mathbf{=}\frac{\mathbf{E}\mathbf{脳}\widehat{\mathbf{n}}}{\mathbf{\left|}\mathbf{E}\mathbf{\right|}}$ 鈥︹�� (3)

(a)

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

${\text{E}}_{\text{0}}=鈭�E_0\cos 45掳\widehat{x}+E_0\sin 45掳\widehat{y}$

Substitute $\frac{{蟽}_0}{{蔚}_0}$for $E_0$into above equation.

$\begin{array}{l}{\text{E}}_0=鈭�\frac{{蟽}_0}{{蔚}_0}\cos 45掳\widehat{x}+\frac{{蟽}_0}{{蔚}_0}\sin 45掳\widehat{y}\\ {\text{E}}_0=鈭�\frac{{蟽}_0}{{蔚}_0}\frac{1}{\sqrt{2}}+\frac{{蟽}_0}{{蔚}_0}\frac{1}{\sqrt{2}}\widehat{y}\\ {\text{E}}_0=\frac{{蟽}_0}{\sqrt{2}{蔚}_0}(鈭�\widehat{x}+纬\widehat{y})\end{array}$

Hence the electrical field$E_0$ in frame$S_0$ is$\frac{{\mathrm{蟽}}_0}{\sqrt{2}{\mathrm{蔚}}_0}(鈭�\widehat{\mathrm{x}}+\mathrm{纬}\widehat{\mathrm{y}})$ .

(b)

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

$\begin{array}{l}{E}_x=E_{x0}\\ E_x=鈭�\frac{{蟽}_0}{\sqrt{2}{蔚}_0}\widehat{x}\end{array}$

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

role="math" localid="1658297161533" $\begin{array}{l}{E}_y=纬E_y\\ E_y=纬\frac{{蟽}_0}{\sqrt{2}{蔚}_0}\widehat{y}\end{array}$

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

$E=E_x+E_y$

Substitute$鈭�\frac{{蟽}_0}{\sqrt{2}{蔚}_0}$ for $E_x$and$纬\frac{{蟽}_0}{\sqrt{2}{蔚}_0}$ for$E_y$ into above equation.

$\begin{array}{l}E=鈭�\frac{{蟽}_0}{\sqrt{2}{蔚}_0}\widehat{x}+纬\frac{{蟽}_0}{\sqrt{2}{蔚}_0}\widehat{y}\\ E=\frac{{蟽}_0}{\sqrt{2}{蔚}_0}(鈭�\widehat{x}+纬\widehat{y})\end{array}$

Hence the electric field$E$ in the frame$S$ is$\frac{{蟽}_0}{\sqrt{2}{蔚}_0}(鈭�\widehat{x}+纬\widehat{y})$ .

(c)

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

Substitute$45掳$ for$胃$ into equation (2).

$\begin{array}{c}\tan 胃=\frac{\sin 45掳}{\frac{1}{纬}\cos 45掳}\\ \tan 胃=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{纬}\frac{1}{\sqrt{2}}}\\ \tan 胃=纬\\ 胃={\tan }^{鈭�1}\left(纬\right)\end{array}$

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

(d)

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

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

$\widehat{n}=鈭�\sin 胃\widehat{x}+\cos 胃\widehat{y}$

Calculate the angle between normal to plates and electric field.

Substitute $\left(\frac{{蟽}_0}{\sqrt{2}{蔚}_0}(鈭�\widehat{x}+纬\widehat{y})\right)$for $E$, $鈭�\sin 胃\widehat{x}+\cos 胃\widehat{y}$ for $\hat{n}$ and $\frac{{蟽}_0}{\sqrt{2}{蔚}_0}\sqrt{1+{纬}^2}$for $\left|E\right|$into equation (3).

$\begin{array}{l}\mathrm{肠辞蝉蠒}=\frac{\left(\frac{{\mathrm{蟽}}_0}{\sqrt{2}{\mathrm{蔚}}_0}(鈭�\widehat{\mathrm{x}}+\mathrm{纬}\widehat{\mathrm{y}})\right)鈰�\left(鈭�\mathrm{蝉颈苍胃}\widehat{\mathrm{x}}+\mathrm{肠辞蝉胃}\widehat{\mathrm{y}}\right)}{\frac{{\mathrm{蟽}}_0}{\sqrt{2}{\mathrm{蔚}}_0}\sqrt{1+{\mathrm{纬}}^2}}\\ \mathrm{肠辞蝉蠒}=\frac{(鈭�\widehat{\mathrm{x}}+\mathrm{纬}\widehat{\mathrm{y}})鈰�(鈭�\mathrm{蝉颈苍胃}\widehat{\mathrm{x}}+\mathrm{肠辞蝉胃}\widehat{\mathrm{y}})}{\sqrt{1+{\mathrm{纬}}^2}}\\ \mathrm{肠辞蝉蠒}=\frac{\mathrm{蝉颈苍胃}+\mathrm{纬肠辞蝉胃}}{\sqrt{1+{\mathrm{纬}}^2}}\\ \mathrm{肠辞蝉蠒}=\frac{\mathrm{肠辞蝉胃}(\mathrm{迟补苍胃}+\mathrm{纬})}{\sqrt{1+{\mathrm{纬}}^2}}\end{array}$ 鈥︹�� (4)

Resolve as:

$\begin{array}{c}\tan 胃=纬\\ \frac{\sin 胃}{\cos 胃}=纬\\ \frac{\sqrt{{\cos }^2胃鈭�1}}{\cos 胃}=纬\\ \sqrt{\frac{1}{{\cos }^2胃}鈭�1}=纬\end{array}$

Solve further as,

$\begin{array}{c}\frac{1}{{\cos }^2胃}鈭�1={纬}^2\\ \frac{1}{{\cos }^2胃}={纬}^2+1\\ {\cos }^2胃=\frac{1}{{纬}^2+1}\\ \cos 胃=\frac{1}{\sqrt{{纬}^2+1}}\end{array}$

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

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

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