Answer to Question #238815 in Electricity and Magnetism for GAYATHRI

Question

3. Among the following which can not be an electric field? E~ = c(xyi +2yzj +3xzk), E~ = c(y^2i + (2xy + z^2)j + 2yzk), where c is a constant.

What would be the dimension of c? (Problem 2.20 of Griffiths, 4th

edition)

Answer 1

Given:

${\bf E}_1 = c(xy{\hat i} +2yz{\hat j} +3xz{\hat k})$

${\bf E}_2 = c(y^2{\hat i} +(2xy+z^2){\hat j} +2yz{\hat k})$

The electrostatic field must satisfy the condition

$\rm rot\:{\bf E}=\bf 0$

Let's check it

$\rm rot\:{\bf E}_1=\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ E_x&E_y&E_z \end{vmatrix}$

$=c\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ xy&2yz&3xz \end{vmatrix}=c(-2y\hat i-3z\hat j-y\hat z)\neq0$

${\rm rot\:{\bf E}_2}=c\begin{vmatrix} \hat i & \hat j &\hat k\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} & \frac{\partial}{\partial z} \\ y^2&2xy+z^2&2yz \end{vmatrix}$

$=c(0\hat i+0\hat j+0\hat z)=\bf 0$

Answer: $E_1$ can not be an electric field.

The dimention of constant $c$ is $\rm V/m^3$