91Ó°ÊÓ

Here, I is the steady current flow between the plates, Ais the area between the plates, dis the distance between the plates, vis the speed of the electrons, $\mathit{ÒÏ}$is the volume charge density.

$\mathit{V}\mathbf{,}\mathit{ÒÏ}\mathbf{,}\mathit{v}$ are all the functions of x plane alone.

a)

The expression for Poisson’s equation for the region between the plates is,

$\frac{{\mathbf{∂}}^{\mathbf{2}}\mathbf{V}}{\mathbf{∂}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{-}{\mathbf{ÒÏ}}_{\mathbf{x}}}{{\mathbf{Î\mu }}_{\mathbf{0}}}$ …… (1)

b)

Write the formula for speed of the electrons using law of conservation of energy.

$$\begin{gathered} \frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}_{\mathbf{x}}^{\mathbf{2}}\mathbf{=}{\mathit{V}}_{\mathbf{x}}\mathit{e} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{v}}_{\mathbf{x}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{r}{\mathbf{V}}_{\mathbf{x}}}{\mathbf{m}}} \end{gathered}$$ …… (2)

Here, ${\mathit{v}}_{\mathbf{x}}\mathit{e}$ is the Potential energy of the electrons, m is the mass.

Therefore, the speed at cathode from rest at point x is ${\mathit{v}}_{\mathbf{x}}\mathbf{=}\sqrt{\frac{\mathbf{2}{\mathbf{rV}}_{\mathbf{x}}}{\mathbf{m}}}$.

c)

Write the expression for volume charge density,

$$\begin{gathered} \mathit{ÒÏ}\mathbf{=}\frac{\mathbf{d}\mathbf{q}}{\mathbf{V}} \\ \mathit{d}\mathit{q}\mathbf{=}\mathit{ÒÏ}\mathit{V} \end{gathered}$$

Here, the value of volumeV can be taken as, $\mathit{a}\mathit{r}\mathit{e}\mathit{a}\mathbf{×}\mathit{s}\mathit{m}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{p}\mathit{o}\mathit{r}\mathit{t}\mathit{i}\mathit{o}\mathit{n}\mathbf{}\mathit{o}\mathit{f}\mathbf{}\mathit{l}\mathit{e}\mathit{a}\mathit{n}\mathit{g}\mathit{h}\mathit{t}\mathbf{=}\mathbf{\left(}\mathbf{Adz}\mathbf{\right)}$

Therefore,

$\mathit{d}\mathit{q}\mathbf{=}\mathit{ÒÏ}\mathbf{\left(}\mathbf{Adx}\mathbf{\right)}$

Write the formula for rate of flow of charge.

$\mathit{I}\mathbf{=}\frac{\mathbf{dq}}{\mathbf{dt}}$ …… (3)

Differentiating the above equation,

$$\begin{gathered} \mathit{I}\mathbf{=}\mathit{A}\mathit{ÒÏ}\frac{\mathbf{dx}}{\mathbf{dt}} \\ \mathbf{}\mathbf{=}\mathit{A}\mathit{ÒÏ}\mathit{v} \end{gathered}$$

Here, current is independent of x.

At steady state I is remains constant.

Thus, as can be seen there is inverse relation between $\mathit{ÒÏ}$ and v.

d)

Using the equation (1), (2) and (3)

$$\begin{gathered} \frac{{\mathbf{∂}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{x}}}{\mathbf{∂}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\frac{\mathbf{-}{\mathbf{ÒÏ}}_{\mathbf{x}}}{{\mathbf{Î\mu }}_{\mathbf{0}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}{\mathbf{I}}_{\mathbf{0}}}{{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{A}{\mathbf{v}}_{\mathbf{x}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{-}{\mathbf{I}}_{\mathbf{0}}}{{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{A}}\sqrt{\frac{\mathbf{m}}{\mathbf{2}\mathbf{e}{\mathbf{V}}_{\mathbf{x}}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{K}}{\sqrt{{\mathbf{V}}_{\mathbf{x}}}} \end{gathered}$$

Here, $\mathit{K}\mathbf{=}\frac{\mathbf{-}{\mathbf{I}}_{\mathbf{0}}}{{\mathbf{Î\mu }}_{\mathbf{0}}\mathbf{A}}\sqrt{\frac{\mathbf{m}}{\mathbf{2}\mathbf{e}}}$is constant.

Thus, $\frac{{\mathbf{∂}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{x}}}{\mathbf{∂}{\mathbf{X}}^{\mathbf{2}}}\mathbf{=}\mathit{K}{\mathit{V}}_{\mathbf{x}}^{\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}}$ …… (4)

Hence, $\frac{{\mathbf{∂}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{x}}}{\mathbf{∂}{\mathbf{X}}^{\mathbf{2}}}\mathbf{=}\mathit{K}{\mathit{V}}_{\mathbf{x}}^{\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}}}$is the required differential equation.

e)

Let’s consider ${\mathit{V}}_{\mathbf{x}}\mathbf{=}\mathit{a}{\mathit{z}}^{\mathbf{b}}$ …… (5)

Differentiate equation (5) twice,

$$\begin{gathered} \frac{\mathbf{d}{\mathbf{V}}_{\mathbf{x}}}{\mathbf{d}\mathbf{x}}\mathbf{=}\mathit{a}\mathit{b}{\mathit{x}}^{\mathbf{b}\mathbf{-}\mathbf{1}} \\ \frac{{\mathbf{d}}^{\mathbf{2}}{\mathbf{V}}_{\mathbf{x}}}{\mathbf{d}{\mathbf{x}}^{\mathbf{2}}}\mathbf{=}\mathit{a}\mathit{b}\left(b-1\right){\mathit{x}}^{\mathbf{b}\mathbf{-}\mathbf{2}} \end{gathered}$$ …… (6)

From the equation (4) and (5)

$$\begin{gathered} \mathit{K}{\mathit{V}}_{\mathbf{x}}^{\mathbf{-}\frac{\mathbf{1}}{\mathbf{2}}}\mathbf{=}\mathit{a}\mathit{b}\left(b-1\right){\mathit{x}}^{\mathbf{b}\mathbf{-}\mathbf{2}} \\ \frac{\mathbf{K}}{\sqrt{\mathbf{a}}}{\mathit{x}}^{\frac{\mathbf{-}\mathbf{b}}{\mathbf{2}}}\mathbf{=}\mathit{a}\mathit{b}\left(b-1\right){\mathit{x}}^{\mathbf{b}\mathbf{-}\mathbf{2}} \end{gathered}$$

Equating the power of x,

role="math" localid="1657269937632" $$\begin{gathered} \mathit{b}\mathbf{-}\mathbf{2}\mathbf{=}\frac{\mathbf{-}\mathbf{b}}{\mathbf{2}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{b}\mathbf{=}\frac{\mathbf{4}}{\mathbf{3}} \end{gathered}$$

After substituting the value of b, equation (5) is written as

${\mathit{V}}_{\mathbf{x}}\mathbf{=}\mathit{a}{\mathit{x}}^{\frac{\mathbf{4}}{\mathbf{3}}}$

At $\mathit{x}\mathbf{=}\mathit{d}$,

$$\begin{gathered} \mathit{V}\mathbf{=}{\mathit{V}}_{\mathbf{0}} \\ {\mathit{V}}_{\mathbf{0}}\mathbf{=}\mathit{a}{\mathit{d}}^{\frac{\mathbf{4}}{\mathbf{3}}} \\ \mathit{a}\mathbf{=}\frac{{\mathbf{V}}_{\mathbf{0}}}{{\mathbf{d}}^{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}}} \end{gathered}$$ …… (7)

Substitute equation (7) in equation (1) and (2).

${\mathit{ÒÏ}}_{\mathbf{x}}\mathbf{=}\mathbf{-}{\mathit{Î\mu }}_{\mathbf{0}}\frac{{\mathbf{V}}_{\mathbf{0}}}{{\mathbf{d}}^{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}}}\left(\frac{4}{9}\right)\frac{\mathbf{1}}{{\mathbf{x}}^{{\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}}}$ …… (8)

${\mathit{V}}_{\mathbf{x}}\mathbf{=}\sqrt{\frac{\mathbf{2}\mathbf{e}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{m}}}\frac{{\mathbf{x}}^{{\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}}}{{\mathbf{d}}^{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}}}$ …… (9)

The above figure shows graph of variation of potential with distance. The dotted line represents as without space charge is linear.

f)

Now combine equation (8) and (9) and solve as further,

$$\begin{gathered} \mathit{I}\mathbf{=}\mathit{A}{\mathit{ÒÏ}}_{\mathbf{x}}{\mathit{V}}_{\mathbf{x}} \\ \mathbf{}\mathbf{=}\mathbf{-}\mathit{A}{\mathit{Î\mu }}_{\mathbf{0}}\frac{{\mathbf{V}}_{\mathbf{0}}}{{\mathbf{d}}^{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}}}\left(\frac{4}{9}\right)\frac{\mathbf{1}}{{\mathbf{x}}^{{\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}}}\sqrt{\frac{\mathbf{2}\mathbf{e}{\mathbf{V}}_{\mathbf{0}}}{\mathbf{m}}}\frac{{\mathbf{x}}^{{\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}}}{{\mathbf{d}}^{{\displaystyle \frac{\mathbf{2}}{\mathbf{3}}}}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathbf{-}\frac{\mathbf{4}\mathbf{A}{\mathbf{Î\mu }}_{\mathbf{0}}}{{\mathbf{d}}^{\mathbf{2}}}\frac{{\mathbf{V}}_{\mathbf{0}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}{\mathbf{9}{\mathbf{x}}^{{\displaystyle \frac{\mathbf{4}}{\mathbf{3}}}}}\sqrt{\frac{\mathbf{2}\mathbf{e}}{\mathbf{m}}} \\ \mathbf{}\mathbf{}\mathbf{=}\mathit{K}{\mathit{V}}_{\mathbf{0}}^{\frac{\mathbf{3}}{\mathbf{2}}} \end{gathered}$$

Here, K is constant and the value is $\mathit{K}\mathbf{=}\frac{\mathbf{-}\mathbf{4}\mathbf{A}}{\mathbf{9}}\sqrt{\frac{\mathbf{2}\mathbf{e}}{\mathbf{m}}}\frac{{\mathbf{Î\mu }}_{\mathbf{0}}}{{\mathbf{d}}^{\mathbf{2}}}$.