SERWAY 24.8
(a) A particle with charge q is located a distance d from an infinite plane. Determine the
electric flux through the plane due to the charged particle.
So, this problem requires you to imagine an infinitely big plane, which is hard.
Look at the picture in diagram (a). Try to imagine the field lines coming off the point
charge. Some of them pass through the blue plane. Now, imagine what happens if you
make the blue plane bigger… it starts to ‘catch’ more field lines.
If we extrapolate this line of reasoning to address an infinitely large plane, I think you
can see that any field line coming off in the downward direction, even if it only comes
off the charge at a very, very, very small downward angle, would be caught by an
infinitely large plane.
The point is this: in the limit that the plane becomes infinitely big, the plane approaches
the limit of catching half the E-field lines that come off the point charge. So, how much
flux does that correspond to?
Imagine surrounding the little point charge with a Gaussian sphere. The infinite plane
would trap exactly half as much of the flux as the Gaussian sphere. But we know what
the flux through a closed Gaussian sphere is. Gauss’s law tells us it only depends on the
amount of charge enclosed…
  q
 Net, Sphere   E  dA  enc
0
So, the net flux through a closed sphere whould be  Net, Sphere 
the infinite plane would therefore be:
 infinite plane 
q
.
2 0
q
0
, and the flux through
(b) What if a particle with charge q is located a very small distance from the center of a
very large square on the line perpendicular to the square and going through the square’s
center?
Look at picture (b) above. Assume the little charge really is a point charge (i.e., it has no
spatial extent). In that case, the point charge could get really close to the plane,
infinitesimally close to the plane, without every touching it.
I think you can see that, in the limit of d going to zero, the finite plane would trap more
and more flux until, again, it approached catching half of the field lines coming off the
charge. So, the answer is the same.
 finite plane, d 0 
q
2 0
(c) Explain why the answers to (a) and (b) are the same.
We’ve already addressed this. The point is, in this situation, the limit as the area of the
plane approaches infinity is mathematically equivalent to the limit as d approaches zero.
In other words, if you put a really big plane really close to a point charge, the point
charge ‘sees’ an infinite plane.