Gauss' Law — Question 5

Question 5

A local linear field reveals charge density through its diagonal gradients

In a small region, $\mathbf E=(Ax+By,Bx+Cy,Dz)$, with constant coefficients. For a small rectangular box, derive the net flux from opposite-face differences rather than assuming the local differential form of Gauss law.

1. Obtain the enclosed volume charge density.

2. For $A=3$, $C=-5$ and $D=4$ in N/(C m), find $\rho/\epsilon_0$. Explain why changing B leaves the inferred density unchanged.

Question 5 - Solution

A local linear field reveals charge density through its diagonal gradients

(a) Method and calculation

$$\Phi_x=A\,\Delta x\Delta y\Delta z,\quad\Phi_y=C\,\Delta x\Delta y\Delta z,\quad\Phi_z=D\,\Delta x\Delta y\Delta z.$$$$\Phi=(A+C+D)\Delta V=\rho\Delta V/\epsilon_0\Rightarrow\rho=\epsilon_0(A+C+D).$$

(b) Interpretation and further calculation

$$\rho/\epsilon_0=3-5+4=2\,\mathrm{N/(C\,m)}.$$$$\partial_x(By)=0,\quad\partial_y(Bx)=0.$$The cross terms contribute equal and opposite flux on opposite faces, although they can change the field and its direction locally.

Physical check

The equal off-diagonal coefficients also give $\partial_xE_y=\partial_yE_x$, consistent with an electrostatic field.