91Ó°ÊÓ

Write the expression for electric filed from divergence theorem

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\frac{\mathbf{ÒÏ}}{{\mathbf{e}}_{\mathbf{0}}}$ …… (1)

Here, $\mathit{E}$is the electric filed, $\mathit{ÒÏ}$is the electric filed, ${\mathit{Î\mu }}_{\mathbf{0}}$is the permittivity for the free space.

Write the formula for electric filed component along with x-axis.

$\mathit{E}\mathbf{=}\mathit{z}\hat{\mathbf{x}}\mathit{x}$ …… (2)

Write the expression for charge density.

$\mathit{ÒÏ}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{[}\mathbf{V}\mathbf{·}\mathbf{E}\mathbf{]}$ …… (3)

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}\mathbf{+}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{y}}}{\mathbf{∂}\mathbf{x}}\mathbf{+}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{2}}}{\mathbf{∂}\mathbf{x}}$ …… (4)

The electric field component along y and z axis is zero. Therefore, $\mathbf{∇}\mathbf{·}\mathit{E}$is expressed as,

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}$ …… (5)

Substitute the value $\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}$for $\mathbf{∇}\mathbf{·}\mathit{E}$in equation (3)

$\mathit{ÒÏ}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{\left[}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}\mathbf{\right]}$

Differentiate ${\mathit{E}}_{\mathbf{x}}$and consider the value from equation (2),

$$\begin{gathered} \frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}\mathbf{=}\frac{\mathbf{∂}}{\mathbf{∂}\mathbf{x}}\mathbf{\left(}\mathbf{ax}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathit{a} \end{gathered}$$

Substitute $\mathit{a}$for $\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}$in equation (3)

$\mathit{ÒÏ}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathit{a}$

From the above equation, it is clear that $\mathit{ÒÏ}$is constant everywhere.

Thus, the charge density is $\mathit{ÒÏ}\mathbf{=}\mathit{C}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$.

Write the expression for charge density by equation (3)

$\mathit{ÒÏ}\mathbf{=}{\mathit{Î\mu }}_{\mathbf{0}}\mathbf{∇}\mathbf{·}\mathit{E}$

From the above equation the charge density is directly proportional to the electric filed.

If charge density is uniform then,

$\mathbf{∇}\mathbf{·}\mathit{E}\mathbf{=}\mathit{C}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}$

Therefore, The values of $\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{x}}}{\mathbf{∂}\mathbf{x}}\mathbf{+}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{y}}}{\mathbf{∂}\mathbf{x}}\mathbf{+}\frac{\mathbf{∂}{\mathbf{E}}_{\mathbf{z}}}{\mathbf{∂}\mathbf{x}}$ are also constant.

Since $\mathit{E}$is linear function of $\mathit{x}\mathbf{,}\mathit{y}$and $\mathit{z}$. From this it is clear that electric field points in particular direction.