91Ó°ÊÓ

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

Charge density of the capacitor plates are $Â\pm {\mathrm{σ}}_0$.

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

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

${\mathbf{E}}_{\mathbf{0}}\mathbf{=}\frac{{\mathbf{σ}}_{\mathbf{0}}}{{\mathbf{Î\mu }}_{\mathbf{0}}}$ …… (1)

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

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

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

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

$\mathbf{³¦´Ç²õÏ•}\mathbf{=}\frac{\mathbf{E}\mathbf{×}\widehat{\mathbf{n}}}{\mathbf{\left|}\mathbf{E}\mathbf{\right|}}$ …… (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 $\frac{{\mathrm{σ}}_0}{{\mathrm{Î\mu }}_0}$for $E_0$into above equation.

$\begin{array}{l}{\text{E}}_0=−\frac{{\mathrm{σ}}_0}{{\mathrm{Î\mu }}_0}\cos 45°\widehat{x}+\frac{{\mathrm{σ}}_0}{{\mathrm{Î\mu }}_0}\sin 45°\widehat{y}\\ {\text{E}}_0=−\frac{{\mathrm{σ}}_0}{{\mathrm{Î\mu }}_0}\frac{1}{\sqrt{2}}+\frac{{\mathrm{σ}}_0}{{\mathrm{Î\mu }}_0}\frac{1}{\sqrt{2}}\widehat{y}\\ {\text{E}}_0=\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}(−\widehat{x}+\mathrm{γ}\widehat{y})\end{array}$

Hence the electrical field$E_0$ in frame$S_0$ is$\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}(−\widehat{\mathrm{x}}+\mathrm{γ}\widehat{\mathrm{y}})$ .

(b)

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

$\begin{array}{l}{E}_x=E_{x0}\\ E_x=−\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}\widehat{x}\end{array}$

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

role="math" localid="1658297161533" $\begin{array}{l}{E}_y=\mathrm{γ}{E}_y\\ E_y=\mathrm{γ}\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}\widehat{y}\end{array}$

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

$E=E_x+E_y$

Substitute$−\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}$ for $E_x$and$\mathrm{γ}\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}$ for$E_y$ into above equation.

$\begin{array}{l}E=−\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}\widehat{x}+\mathrm{γ}\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}\widehat{y}\\ E=\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}(−\widehat{x}+\mathrm{γ}\widehat{y})\end{array}$

Hence the electric field$E$ in the frame$S$ is$\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}(−\widehat{x}+\mathrm{γ}\widehat{y})$ .

(c)

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

Substitute$45°$ for$\mathrm{θ}$ into equation (2).

$\begin{array}{c}\tan \mathrm{θ}=\frac{\sin 45°}{\frac{1}{\mathrm{γ}}\cos 45°}\\ \tan \mathrm{θ}=\frac{\frac{1}{\sqrt{2}}}{\frac{1}{\mathrm{γ}}\frac{1}{\sqrt{2}}}\\ \tan \mathrm{θ}=\mathrm{γ}\\ \mathrm{θ}={\tan }^{−1}\left(\mathrm{γ}\right)\end{array}$

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

(d)

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

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

$\widehat{n}=−\sin \mathrm{θ}\widehat{x}+\cos \mathrm{θ}\widehat{y}$

Calculate the angle between normal to plates and electric field.

Substitute $\left(\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}(−\widehat{x}+\mathrm{γ}\widehat{y})\right)$for $E$, $−\sin \mathrm{θ}\widehat{x}+\cos \mathrm{θ}\widehat{y}$ for $\hat{n}$ and $\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}\sqrt{1+{\mathrm{γ}}^2}$for $\left|E\right|$into equation (3).

$\begin{array}{l}\mathrm{³¦´Ç²õÏ•}=\frac{\left(\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}(−\widehat{\mathrm{x}}+\mathrm{γ}\widehat{\mathrm{y}})\right)â‹\dots \left(−\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{x}}+\mathrm{³¦´Ç²õθ}\widehat{\mathrm{y}}\right)}{\frac{{\mathrm{σ}}_0}{\sqrt{2}{\mathrm{Î\mu }}_0}\sqrt{1+{\mathrm{γ}}^2}}\\ \mathrm{³¦´Ç²õÏ•}=\frac{(−\widehat{\mathrm{x}}+\mathrm{γ}\widehat{\mathrm{y}})â‹\dots (−\mathrm{²õ¾\pm ²Ôθ}\widehat{\mathrm{x}}+\mathrm{³¦´Ç²õθ}\widehat{\mathrm{y}})}{\sqrt{1+{\mathrm{γ}}^2}}\\ \mathrm{³¦´Ç²õÏ•}=\frac{\mathrm{²õ¾\pm ²Ôθ}+\mathrm{㳦´Ç²õθ}}{\sqrt{1+{\mathrm{γ}}^2}}\\ \mathrm{³¦´Ç²õÏ•}=\frac{\mathrm{³¦´Ç²õθ}(\mathrm{³Ù²¹²Ôθ}+\mathrm{γ})}{\sqrt{1+{\mathrm{γ}}^2}}\end{array}$ …… (4)

Resolve as:

$\begin{array}{c}\tan \mathrm{θ}=\mathrm{γ}\\ \frac{\sin \mathrm{θ}}{\cos \mathrm{θ}}=\mathrm{γ}\\ \frac{\sqrt{{\cos }^2\mathrm{θ}−1}}{\cos \mathrm{θ}}=\mathrm{γ}\\ \sqrt{\frac{1}{{\cos }^2\mathrm{θ}}−1}=\mathrm{γ}\end{array}$

Solve further as,

$\begin{array}{c}\frac{1}{{\cos }^2\mathrm{θ}}−1={\mathrm{γ}}^2\\ \frac{1}{{\cos }^2\mathrm{θ}}={\mathrm{γ}}^2+1\\ {\cos }^2\mathrm{θ}=\frac{1}{{\mathrm{γ}}^2+1}\\ \cos \mathrm{θ}=\frac{1}{\sqrt{{\mathrm{γ}}^2+1}}\end{array}$

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

$\begin{array}{l}\cos \mathrm{ϕ}=\frac{1}{\sqrt{1+{\mathrm{γ}}^2}}\frac{(\mathrm{γ}+\mathrm{γ})}{\sqrt{1+{\mathrm{γ}}^2}}\\ \cos \mathrm{ϕ}=\frac{2\mathrm{γ}}{1+{\mathrm{γ}}^2}\end{array}$

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