Suppose an electric field$\mathit{E}\left(x,y,z\right)$ has the form

role="math" localid="1657526371205" ${\mathit{E}}_{\mathbf{x}}\mathbf{=}\mathit{a}\mathit{x}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}{\mathit{E}}_{\mathbf{y}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{}\mathbf{}\mathbf{}\mathbf{}{\mathit{E}}_{\mathbf{z}}\mathbf{=}\mathbf{0}$

Where $\mathit{a}$is a constant. What is the charge density? How do you account for the fact that the field points in a particular direction, when the charge density is uniform?

Write the expression for electric filed from divergence theorem

$\mathbf{\nabla }\mathbf{.}\mathit{E}\mathbf{=}\frac{\mathbf{\rho }}{{\mathbf{\epsilon }}_{\mathbf{0}}}$ …… (1)

Here,$E$ is the electric filed, $\rho$is the electric filed,${\epsilon }_0$ is the permittivity for the free space.

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

$E=ax\stackrel{⏜}{x}$ …… (2)

Write the expression for charge density.

$\rho ={\epsilon }_0\left[\nabla .E\right]$ …… (3)

$\nabla .E=\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial x}+\frac{\partial E_z}{\partial x}$ …… (4)

The electric field component along y and z axis is zero. Therefore,$\nabla .E$is expressed as,

$\nabla .E=\frac{\partial E_x}{\partial x}$ …… (5)

Substitute the value$\frac{\partial E_x}{\partial x}\mathrm{for}\nabla .E$in equation (3)

$\rho ={\epsilon }_0=\left[\frac{\partial E_x}{\partial x}\right]$

Differentiate $E_x$and consider the value from equation (2),

$$\begin{gathered} \frac{\partial E_x}{\partial x}=\frac{\partial }{\partial x}\left(ax\right) \\ =a \end{gathered}$$

Substitute for$\frac{\partial E_x}{\partial x}$ in equation (3)

$\rho ={\epsilon }_{0}a$

From the above equation, it is clear that$\rho$ is constant everywhere.

Thus, the charge density is$\rho =\mathrm{Constant}$ .

Write the expression for charge density by equation (3)

$\rho ={\epsilon }_0\nabla .E$

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

If charge density is uniform then,

$\nabla .E=Cons\tan t$

Therefore, The values of $\frac{\partial E_x}{\partial x}+\frac{\partial E_y}{\partial x}+\frac{\partial E_z}{\partial x}$are also constant.

Since $E$is linear function ofrole="math" localid="1657527677785" $x,y\mathrm{and}\mathrm{z}$. From this it is clear that electric field points in particular direction.