Q2.55P - Introduction to Electrodynamics

Question

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

                               ${\mathit{E}}_{\mathbf{x}}\mathbf{=}\mathit{a}\mathit{x}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}_{\mathbf{y}}\mathbf{=}\mathbf{0}\mathbf{,}\mathbf{ }\mathbf{ }\mathbf{ }\mathbf{ }{\mathit{E}}_{\mathbf{z}}\mathbf{=}\mathbf{0}$

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?

Step-by-Step Solution

Answer

Answer

The formula of charge density is $\mathit{\rho }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\left[\nabla ·E\right]$.

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

Step 1: Define functions

Write the expression for electric filed from divergence theorem

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

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

Step 2: Determine charge density

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.

$\mathit{\rho }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{[}\mathbf{V}\mathbf{·}\mathbf{E}\mathbf{]}$                               …… (3)

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

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

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

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

$\mathit{\rho }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\left[}\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{x}}}{\mathbf{\partial }\mathbf{x}}\mathbf{\right]}$

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

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

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

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

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

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

Step 3: Determine charge density is uniform

Write the expression for charge density by equation (3)

$\mathit{\rho }\mathbf{=}{\mathit{\epsilon }}_{\mathbf{0}}\mathbf{\nabla }\mathbf{·}\mathit{E}$           

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

If charge density is uniform then,         

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

Therefore, The values of $\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{x}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{y}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\frac{\mathbf{\partial }{\mathbf{E}}_{\mathbf{z}}}{\mathbf{\partial }\mathbf{x}}$  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.