An infinitely large thin sheet in the \(xy\)-plane carries uniform positive charge density and is moving with constant velocity \(\vec{v}\) in the \(+x\) direction (see figure below). The direction of the corresponding Poynting vector is
Show Hint
Use \(E=\sigma/2\epsilon_0\) away from the sheet and \(B=\mu_0 K/2\) (with \(K=\sigma v\)) circling the current sheet like a solenoid winding; both \(E\) and \(B\) flip sign together across the sheet, so \(\vec S = \vec E \times \vec B/\mu_0\) keeps the same sign on both sides.
Step 1: Set up the fields with a general height coordinate. Let $h$ stand for the field point's $z$-coordinate, positive or negative. Both fields on either side of the sheet can be written compactly using the sign function $\text{sgn}(h)$: \[ \vec E = \frac{\sigma}{2\epsilon_0}\,\text{sgn}(h)\,\hat z, \qquad \vec B = -\frac{\mu_0\sigma v}{2}\,\text{sgn}(h)\,\hat y \] This single-formula form already shows that $\vec E$ and $\vec B$ each pick up exactly one factor of $\text{sgn}(h)$ when you cross the sheet.
Step 2: Take the cross product symbolically before plugging in signs. \[ \vec S = \frac{1}{\mu_0}\vec E \times \vec B = \frac{1}{\mu_0}\left(\frac{\sigma}{2\epsilon_0}\text{sgn}(h)\hat z\right)\times\left(-\frac{\mu_0\sigma v}{2}\text{sgn}(h)\hat y\right) \] \[ \vec S = -\frac{\sigma^2 v}{4\epsilon_0}\big[\text{sgn}(h)\big]^2 (\hat z \times \hat y) \]
Step 3: Use $[\text{sgn}(h)]^2 = 1$. Whatever the sign of $h$, squaring it always gives $1$, since $\text{sgn}(h) = \pm 1$. That single fact already tells you the final direction cannot depend on whether $z$ is positive or negative, before even working out the cross product of the unit vectors. This immediately rules out options (B), (C) and (D), which all claim the direction is different on the two sides.
Step 4: Work out the fixed direction. With $\hat z \times \hat y = -\hat x$: \[ \vec S = -\frac{\sigma^2 v}{4\epsilon_0}(1)(-\hat x) = \frac{\sigma^2 v}{4\epsilon_0}\hat x \] which is along $+x$, matching the direction the sheet itself is moving in.
Step 5: Physical sanity check. This makes sense energetically: a moving charged sheet is effectively carrying electromagnetic energy along with it as it travels, and the electromagnetic momentum density $\vec S/c^2$ should point the same way the charge is physically moving, on both faces of the sheet. A field pattern that reversed on the two sides would not match this simple picture of energy being dragged along by the moving charge.
Final Answer: The Poynting vector is $+x$ for both $z<0$ and $z>0$.\[ \boxed{\text{(A)}} \]
