Question

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

• A. \(+x\) for both \(z < 0\) and \(z > 0\)
• B. \(+x\) for \(z < 0\) and \(-x\) for \(z > 0\)
• C. \(-x\) for \(z < 0\) and \(+x\) for \(z > 0\)
• D. \(-x\) for both \(z < 0\) and \(z > 0\)

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.

Answer 1

The Correct Option is A

Step 1: Find the electric field of the sheet.
An infinite sheet with uniform positive surface charge density \(\sigma\) sitting in the \(xy\)-plane sets up a uniform electric field pointing straight away from the sheet on each side, of magnitude \(\sigma/(2\epsilon_0)\). So
\[ \vec{E} = \frac{\sigma}{2\epsilon_0}\hat{z} \ \text{for } z>0, \qquad \vec{E} = -\frac{\sigma}{2\epsilon_0}\hat{z} \ \text{for } z<0 \]

Step 2: Find the magnetic field of the moving sheet.
A charged sheet moving at velocity \(\vec{v}=v\hat{x}\) is a surface current \(K = \sigma v\) flowing in the \(+x\) direction. Building this field by adding up the fields of parallel line currents (each an infinite wire along \(x\)) and integrating across the sheet gives, on each side, a field along \(\hat y\) whose sign flips across the sheet, exactly like the field of a solenoid winding:
\[ \vec{B} = -\frac{\mu_0 \sigma v}{2}\hat{y} \ \text{for } z>0, \qquad \vec{B} = \frac{\mu_0 \sigma v}{2}\hat{y} \ \text{for } z<0 \]

Step 3: Compute the Poynting vector for \(z>0\).
The Poynting vector is \(\vec{S} = \frac{1}{\mu_0}\vec{E}\times\vec{B}\). Using \(\hat z \times \hat y = -\hat x\):
\[ \vec{S} = \frac{1}{\mu_0}\left(\frac{\sigma}{2\epsilon_0}\hat z\right)\times\left(-\frac{\mu_0\sigma v}{2}\hat y\right) = -\frac{\sigma^2 v}{4\epsilon_0}(\hat z \times \hat y) = \frac{\sigma^2 v}{4\epsilon_0}\hat x \]
So the Poynting vector points along \(+x\) for \(z>0\).

Step 4: Compute the Poynting vector for \(z<0\).
Here both \(\vec E\) and \(\vec B\) flip sign compared with the \(z>0\) side, and two sign flips cancel:
\[ \vec{S} = \frac{1}{\mu_0}\left(-\frac{\sigma}{2\epsilon_0}\hat z\right)\times\left(\frac{\mu_0\sigma v}{2}\hat y\right) = -\frac{\sigma^2 v}{4\epsilon_0}(\hat z \times \hat y) = \frac{\sigma^2 v}{4\epsilon_0}\hat x \]
which again points along \(+x\). So the same-sign flip in both \(E\) and \(B\) keeps the energy flow direction fixed on both sides of the sheet.

Step 5: Why the other options are wrong.
Options (B) and (D) both claim the Poynting vector flips sign across the sheet, but since \(\vec E\) and \(\vec B\) both flip sign together when crossing from \(z>0\) to \(z<0\), their cross product does not flip, so the flow direction stays the same on both sides. Option (C) claims the field points in \(-x\), which is the opposite of what the cross product \(\hat z \times (-\hat y) = \hat x\) actually gives.

Final Answer:
The Poynting vector points in the \(+x\) direction on both sides of the sheet, in the same sense as the sheet's own motion.\[ \boxed{\text{(A)}} \]