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Solutions for Friday Homework 3, Feb 01, 2008
22.32) Start with the definition of electric flux:
φe =
E^ " · d A"
So far we have mostly been thinking of the dot product as
A^ " · B" = |A| |B| cos θ
but we can also use
A^ " · B" = Ax Bx + Ay By + Az Bz
where Ax is the component of A" lying in the x direction, and so on. Recall that for any vector A",
A^ " = Ax^ ˆi + Ay ˆj + Az ˆk
where ˆi is a vector of length=1 (a “unit vector”) pointing in the +x direction, and likewise for the others.
We can use this form of the dot product to rewrite the integral:
φe =
(Ex · dAx + Ey · dAy + Ez · dAz )
Ex dAx +
Ey dAy +
Ez dAz
The problem indicates that the electric field is constant, therefore each component of the field is constant. Since constants can be pulled out of the integral:
φe = Ex
dAx + Ey
dAy + Ez
dAz
Then since
d(anything) = anything this becomes:
φe = Ex Ax + Ey Ay + Ez Az
We are given the components of E", so we just need to figure out what the components of A" are on each face. Let’s start with s 1 , the face to the right in the picture. The magnitude of an area vector is just the area it represents, so this vector will have length L 2. The direction of an area vector is perpendicular to the surface, pointing out from the enclosed volume. So for s 3 the area vector points in the negative y direction, so
A^ " 1 = 0ˆi − L 2 ˆj + 0ˆk
and so
φ 1 = (−B)(0) + C(−L 2 ) + (−D)0 = −CL 2
Similarly for the other sides,
φ 2 = (−D)(L 2 ) = −DL 2 φ 3 = (C)(L 2 ) = CL 2 φ 4 = (−D)(−L 2 ) = DL 2 φ 5 = (−B)(L 2 ) = −BL 2 φ 6 = (−B)(−L 2 ) = BL 2
b) Adding together the fluxes from each face gives the flux through the whole cube:
φ =
φi = L 2 (−C − D + C + D − B + B) = 0
Since the electric field is constant every field line that goes into the cube must also come out, so we might expect that the total flux would be zero.
By symmetry the electric field E" must point outward. At the top the area vector points up – the two vectors are perpendicular so the flux through the top is zero. Likewise for the bottom.
Along the side of the cylinder the area vector always points radially outward, in the same direction as the electric field. So the angle between the area vector and the field vector is 0, so
φside =
E^ " · d A" =
E cos(0)dA =
EdA
Again by symmetry E" is constant, so
φside = E
dA = EA
To get the surface area of the side, imagine the cylinder is a roll-up piece of paper. When it is unrolled the base will be the length of the circumference, so
A = (2πr)h
Then putting the pieces together:
πr 2 hρ $ (^0)
= (2πr)hE
Note that the factors of h cancel. This is a good sign, since there is nothing physical about h, it is just an arbitrary height we selected for our “imaginary” Gaussian cylinder and so the result should not depend on it.
Solving for E:
E =
ρ 2 $ (^0)
r
Note that this increases linearly with r. Compare this to the case of a solid sphere of charge discussed in lecture!
b) For the outside we’ll go through the same procedure. For the Gaussian surface we’ll use a cylinder outside the physical cylinder.
Then Gauss’ law says:
φe =
E^ " · d A" = Q^ enc $ (^0)
The integral on the right is exactly the same as in part (a):
φe = EA = (2πr)hE
Now the amount of charge enclosed is just a function of the height,
Q (^) enc = hλ
so putting these together and solving for E,
E =
λ 2 π$ (^0)
r
Note this is identical to the case of a wire (see the last equation on example 21.11, page 731). This is just like how outside a sphere of charge it resembles a point charge.
