91Ó°ÊÓ

The wis the width of ribbons.

The h is the separation between two ribbons.

The I is the length of the ribbon.

Write the formula ofcapacitance per unit length.

$\mathbf{C}\mathbf{=}\frac{\mathbf{Q}}{\mathbf{V}}$ …… (1)

Here, C is the capacitance and V is the voltage.

Write the formula ofinductance per unit length.

$\mathit{ϕ}\mathbf{=}\mathbf{LI}$ …… (2)

Here, L is inductance per unit length and I is the length of the ribbon.

Write the formula ofpropagation speed.

$\mathbf{v}\mathbf{=}\frac{\mathbf{1}}{\sqrt{{\mathbf{μ}}_{\mathbf{0}}{\mathbf{Î\mu }}_{\mathbf{0}}}}$ …… (3)

Here,${\mathit{Î\mu }}_{\mathbf{0}}$ is permittivity and${\mathit{μ}}_{\mathbf{0}}$ is permeability.

Write the formula ofpropagation speed.

$\mathbf{L}\mathbf{=}\frac{\mathbf{μ³ó}}{\mathbf{w}}$ …… (4)

Here,$\mathit{μ}$ is permeability, h is the separation between two ribbons and w is the width of ribbons.

Now we discuss for a parallel plate capacitor.

The electric field between the plates of the capacitor is

$E=\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}$

The potential difference between the plates of capacitor is

$V=Eh$

Then $V=\left(\frac{\mathrm{σ}}{{\mathrm{Î\mu }}_0}\right)h$

But role="math" localid="1657533956022" $\mathrm{σ}=\mathrm{Charge}\mathrm{density}=\frac{Q}{wI}$

Then $V=\frac{1}{{\mathrm{Î\mu }}_0}\left(\frac{Q}{wI}\right)h$

Determine the capacitance per unit length.

Substitute$\frac{1}{{\mathrm{Î\mu }}_0}\left(\frac{Q}{wI}\right)h$ for v into equation (1).

$$\begin{gathered} C=\frac{Q}{{\displaystyle \frac{1}{{\mathrm{Î\mu }}_0}}\left({\displaystyle \frac{Q}{wI}}\right)h} \\ =\frac{{\mathrm{Î\mu }}_{0}wI}{h} \end{gathered}$$

Therefore, the value of capacitance per unit length is$\mathrm{C}=\frac{{\mathrm{Î\mu }}_0\mathrm{w}\mathrm{I}}{\mathrm{h}}$ .

We know that magnetic field.

$$\begin{gathered} B={\mathrm{μ}}_{0}k \\ k=\frac{I}{w} \end{gathered}$$

Then $$\begin{gathered} B=\frac{{\mathrm{μ}}_{0}I}{w} \\ \mathrm{ϕ}=BhI \end{gathered}$$

Then $=\frac{{\mathrm{μ}}_{0}I}{w}hL$

Determine inductance per unit length.

Substitute$\frac{{\mathrm{μ}}_{0}IhI}{w}$ for$\mathrm{ϕ}$ into equation (2).

$$\begin{gathered} \frac{{\mathrm{μ}}_{0}IhI}{w}=LI \\ L=\frac{{\mathrm{μ}}_{0}IhI}{w} \end{gathered}$$

Then inductance per unit length.

$$\begin{gathered} \frac{L}{I}=L \\ =\frac{{\mathrm{μ}}_{0}h}{w} \end{gathered}$$

Therefore, the value of inductance per unit length is $L=\frac{{\mathrm{μ}}_{0}h}{w}$.

Derive the propagation speed as follows:

$$\begin{gathered} LC=\frac{{\mathrm{Î\mu }}_{0}w}{h}×\frac{{\mathrm{μ}}_{0}h}{w} \\ ={\mathrm{μ}}_0{\mathrm{Î\mu }}_0 \\ =\left(4\mathrm{Ï€}×{10}^{-7}\mathrm{H}/\mathrm{m}\right)\left(8.85×{10}^{-12}{\mathrm{C}}^2/{\mathrm{Nm}}^2\right) \\ =1.112×{10}^{-17}{\mathrm{s}}^2/{\mathrm{m}}^2 \end{gathered}$$

Determine the propagation speed.

Substitute $\mathrm{LC}$for$\sqrt{{\mathrm{μ}}_0{\mathrm{Î\mu }}_0}$ into equation (3).

$$\begin{gathered} v=\frac{1}{\sqrt{\mathrm{LC}}} \\ =2.999×{10}^8\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the value of propagation speed $v=2.99×{10}^8\mathrm{m}/\mathrm{s}$.

As a non-conducting substance with permittivity $\mathrm{Î\mu }$and permeability $\mathrm{μ}$separates the strips from one another as follows:

$$\begin{gathered} D=\mathrm{σ} \\ E=\frac{D}{\mathrm{Î\mu }} \\ E=\frac{\mathrm{σ}}{\mathrm{Î\mu }} \end{gathered}$$

Then solve as:

$$\begin{gathered} C=\frac{\mathrm{Î\mu }w}{h} \\ H=K \\ B=\mathrm{μ}H \\ B=\mathrm{μ}K \end{gathered}$$

Determine propagation speed.

Substitute$\mathrm{Î\mu }$for$\frac{h}{w}$into equation (4).

localid="1657535319966" $$\begin{gathered} LC=\mathrm{Î\mu }\mathrm{μ} \\ v=\frac{1}{\sqrt{\mathrm{Î\mu }\mathrm{μ}}} \end{gathered}$$

Therefore, the value of propagation speed $v=\frac{1}{\sqrt{\mathrm{Î\mu }\mathrm{μ}}}$.