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An infinitely long uniform line charge (charge per unit length = $\lambda$) lies parallel to the $y$-axis in the $y$-$z$ plane at $z=\frac{\sqrt{3}a}{2}$. Let $ABCD$ be a rectangle in the $x$-$y$ plane with origin as centre and with sides parallel to the $x$ and $y$ axes (Length of the side along $y$-axis = $L$ and Length of side along $x$-axis = $a$). Find the magnitude of total flux through $ABCD$. enter image description here

So what I've tried (and seen online) is to make a regular hexagonal prism around the line charge with $ABCD$ as one of the rectangular sides and then apply Gauss Law. The online answers simply say the flux is $\frac{\lambda L}{6\epsilon}$ as the total flux $\frac{\lambda L}{\epsilon}$ is evenly spaced between the $6$ rectangular sides. However we aren't accounting for the flux through the hexagonal planes. How do we explain this? Any clarity would be appreciated.

Here is a link to one of those online answers.

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By taking the line of charge to be infinite, we've introduced the fact that translation along the y axis is invariant; if we measure the field around the wire at some point, then move along the wire, we expect to see the same field as our system still looks the same, i.e. we have no way of measuring our 'y-coordinate' along the wire. Now by symmetry, if we turn around, the wire still looks the same; we expect the field to be identical then. Now let's suppose we introduce some y-dependence on the field, so it points in a direction. This breaks the symmetry, as we can now define a 'positive' y-direction. So we need the field to be purely radial, and hence no component of the field points perpendicular to the hexagonal plane. Hence the flux through these planes is $0$.

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  • $\begingroup$ It would be nicer if you clarify the hexagonal plane you are talking about. Because it is harder to comprehend without any indication of what that plane is. $\endgroup$ – FakeMod Apr 3 at 7:23
  • $\begingroup$ So as far as I understand this question, the hexagonal planes are the two planes bounding the surface with normal vectors parallel to the y-axis; hence there is no flux through either. $\endgroup$ – EigenFunction Apr 3 at 7:24
  • $\begingroup$ @EigenFunction Exactly! But I am just saying that your answer will be improved if you add an image showing those planes, like this $\endgroup$ – FakeMod Apr 3 at 7:31
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Short answer: There is no flux through the hexagonal faces.

Long answer: It starts with the fact that at any point in space, the electric field of the line charge (LC) is directed radially outward from the wire (in direction perpendicular to the length of the LC) To see why, we can consider a symmetrical argument:-

https://i.stack.imgur.com/UotfW.png

As you can see, for every point A on the line charge, there exists a point B on the line charge such that the fields at P due to the two point charges A and B have equal and opposite vertical components. Hence, the resultant fields of all such pairs of points is purely horizontal. (Here, horizontal and vertical are with respect to the figure).

Since the hexagonal faces are also perpendicular to the wire, the area vector of the hexagonal faces is perpendicular to the electric field. Hence,

$\phi = ES\cos\theta = 0 $

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