SERWAY 24.27
Consider a long, cylindrical charge distribution of radius R with a uniform charge density
ρ. Find the electric field at distance r from the axis, where r < R.
I did this problem in class, so I’m not going to give an exhaustive answer here. That
said, let’s write down Gauss’s Law:
  q
 E   E  dA  enc
0
We’re dealing with a cylindrically symmetrical, uniform charge distribution that is
essentially infinitely long. Thus, the Gaussian surface of choice, the blue cylinder,
exhibits the same symmetry and intersects the point in space where we’re interested in
knowing the E-field (a distance r away from the charge distribution’s central axis).
Realize, the net E-flux through the Gaussian surface has contributions from the flux
through the base, the flux through the top, and the flux through the sides. In other
words:
 E   top   base   sides
Put yourself on the top of the blue Gaussian surface. The dA vector there points to the
left, along the axis of the charge distribution. The E-field, on the other hand, must point
radially outward from the distribution’s central axis (because all left-right components
of the E-field cancel out, since this charge distribution is infinitely long). So, no matter
where you stand on the top of the blue cylinder, the dA vectors and the E-field vectors
are always perpendicular. Thus, the E-field doesn’t ‘push through’ the top of the blue
cylinder at all, and its contribution to the net flux is simply zero. The same analysis
holds for the base of the blue cylinder.
So, Gauss’s Law simplifies because:
  q
 E   top   base   sides  0  0   sides   E  dA  enc
0
The only area we really care about is just the sides of the blue cylinder.
Correspondingly, the closed surface integral in Gauss’s law simplifies to:
  q
 E   sides   E  dA  enc
0
sides
Well, put yourself on the sides of the blue cylinder. No matter where you stand on its
surface, the dA vectors point radially outward. Also, no matter where you stand on its
sides, the E-field vectors point radially outward (again, the left-right components of the
E-field cancel since the charge distribution is infinitely long). So, the angle between the
E-field and the dA vectors is always zero:
 E   EdA cos  
sides
 E   EdA 
qenc
0
qenc
sides
0
Now, because the infinitely long charge distribution always looks the same to you, no
matter where you stand on the sides of the blue cylinder, then the magnitude of the Efield must be constant at all points on the sides of the cylinder. Thus:
E  dA 
sides
qenc
0
But  dA just represents the surface area of the sides of the blue cylinder, which we
sides
know from our studies of geometry is simply 2  rl  , where l represents the length of
the blue Gaussian cylinder. Plugging this result back in…
E 2  rl  
qenc
0
Now, we must determine how much charge in actually enclosed in the Gaussian surface.
Well, the charge distribution has a uniform volumetric charge density, ρ. So, the
amount of charge inside the Gaussian cylinder is just the density of the charge
multiplied by the volume encompassing the charge:

q enc    VGaussian      r 2 l

Thus, Gauss’s Law becomes:
  r
E 2  rl  
2
l
0
Canceling like terms and rearranging, we get:
 r
, and the field points radially away from the central axis of the charge
2 0
distribution.
E