Charged Dome Problem
(a) A very thin, non-conducting, hemispherical dome of radius R is charged uniformly
with total charge +Q. It lies with its open downward face centered at the origin in the x-y
plane.
Determine both the direction and magnitude of the electric field E at the center point of
the hemisphere, (x,y) = (0,0).
Solution to (a)
(a) First, notice that there is no easy way to solve this problem using Gauss’s Law.
Gauss’s Law is true here, it just isn’t useful. That’s because the electric field is not
constant on any surface of known area containing the origin, so there just isn’t any S for
which knowing ∯𝑆 𝐸⃗ ∙ 𝑑𝑎 is of any particular use to us.
𝑘𝑑𝑞
Instead, we can use Coulomb’s Law (𝑑𝐸⃗ = 2 𝑟̂ ) to add up the contributions to the field
𝑟
from all the little patches on the sphere. Now, it seems that we have a problem, because
we want to integrate, but each little patch of sphere is going to contribute a field with a
different direction! From calculus we know how to add up a bunch of tiny numbers, but
what about a bunch of tiny arrows?
One solution is to break each arrow up into three numbers (‘components’), and then add
up the three resulting lists of tiny numbers separately. At the end, the three totals can be
put back together to form the final arrow.
However, here we can take a bit of a shortcut. Notice that we could rotate this whole
dome through some angle φ about the vertical axis, and the problem wouldn’t change.
We typically express this by saying that there is ‘azimuthal symmetry’. (Although the
problem involves part of a sphere, there is not spherical symmetry in this problem.
Spherical symmetry would mean we could also rotate the dome through an angle θ about
a horizontal axis without changing the problem, but we can’t.) The fact that we have this
symmetry means that the answer to our problem cannot depend on the angle φ. In
particular, there can’t be any component of the field perpendicular to the z –axis; if there
were, we could rotate the dome about the z-axis and get a different answer, but we said
that wasn’t acceptable because the dome is the same no matter what φ is.
(This might seem a bit abstract, but if you do enough problems like this you’ll probably end up
liking this style of reasoning, because it is concise and powerful. In fact, you could even
summarize the whole argument as “by azimuthal symmetry, the field has to be along the z-axis.”
If that does seem too abstract for now though, you could also think of it like this: the field has to
be in the z-direction because for each patch of the sphere in one direction, there is also a patch of
sphere “opposite”, i.e. rotated 180˚ about the z-axis. The components of these two fields which
are perpendicular to the z-axis will be equal and opposite each other (why?), and will cancel.)
Alright, so we can ignore everything but the z-components of the field. That’s two thirds
of the problem (x and y components both zero). Now all we have to do is write down the
z-component of the field due to each little patch of the sphere, and add up the fields from
all the little patches.
𝑑𝑄 = 𝜎𝑑𝐴 = 𝜎𝑅 2 sin 𝜃 𝑑𝜃𝑑𝜑,
where 𝜎 =
𝑄
2𝜋𝑅2
is the surface charge density, and 𝑑𝐴 is the area of the small patch.
Why does 𝑑𝐴 = 𝑅 2 sin 𝜃 𝑑𝜃𝑑𝜑? You have two options: (a) it just does; (b) because 𝑅 2 sin 𝜃 is
the determinant of the matrix of partial derivatives, i.e. the Jacobian for transforming from
Cartesian to spherical coordinates. If you aren’t satisfied with (a) but (b) doesn’t sound familiar
from calculus class, you should come to my office hours for a more detailed answer. Intuitively,
it has to depend on R2 because the length and width of the little patch both scale linearly with R;
and you need the sinθ because when you rotate about the z-axis from (θ,φ) to (θ,φ+dφ), your
‘effective radius’ (don’t read too much into that, I’m just “handwaving” for brevity) is only
Rsinθ.
Now the field due to this small patch is:
𝑑𝐸⃗ = −
̂
𝑑𝑄𝑹
4𝜋𝜀 ̥𝑅
=−
2
̂
𝜎𝑅 2 𝑠𝑖𝑛 𝜃 𝑑𝜃𝑑𝜑𝑹
4𝜋𝜀 ̥𝑅2
=−
̂
𝜎𝑠𝑖𝑛 𝜃 𝑑𝜃𝑑𝜑𝑹
4𝜋𝜀 ̥
,
̂ is
where the first equality is from Coulomb’s law, then just subbing in and simplifying. 𝑹
the unit vector in the direction of the little patch whose field we’re calculating; i.e. it
points from the origin toward that little patch. The minus sign is because the field points
away from the patch.
However, we don’t want the whole field due to each patch, we just want the zcomponent. (Remember, we know how to add numbers, not arrows.) Thus,
𝜎𝑠𝑖𝑛 𝜃 cos 𝜃𝑑𝜃𝑑𝜑
𝑑𝐸𝑧 = 𝑑𝐸⃗ ∙ 𝑧̂ = −|𝑑𝐸⃗ |𝑐𝑜𝑠𝜃 = −
.
4𝜋𝜀 ̥
Why can we write the magnitude of the z-component this way? You’ll probably need to
̂ is
look at a picture to convince yourself, but the angle between the negative z-axis and 𝑹
θ, and we want the parallel component, so we use cosine.
Now we just need to integrate this over θ and φ:
2𝜋
𝜋
2
𝐸𝑧 = ∫ 𝑑𝐸𝑧 = − ∫ 𝑑φ ∫
𝜋
2
= −2𝜋
0
0
𝜎𝑠𝑖𝑛 𝜃 cos 𝜃 𝑑𝜃
4𝜋𝜀 ̥
𝜎
∫ 𝑠𝑖𝑛𝜃𝑐𝑜𝑠𝜃𝑑𝜃
4𝜋𝜀 ̥
0
𝜋
2
=
𝜎
1
∫ 𝑠𝑖𝑛2𝜃𝑑𝜃
2𝜀 ̥ 2
0
=
𝜎
1
𝜋/2
[− 𝑐𝑜𝑠2𝜃]
0
4𝜀 ̥ 2
𝐸𝑧 =
𝜎
𝑄
=
4𝜀 ̥ 8𝜋𝜀 ̥𝑅 2
And that’s the answer! The field has magnitude
𝜎
, and it points in the −𝑧̂ direction.
4𝜀 ̥
Charged Dome, Part (b)
(b) What is the flux of E produced through the x-y plane by this charged hemisphere?
Show your work or discuss how you find this result. (Hint: Sketch lines of E produced by
the charged hemisphere. Think about what Gaussian surfaces you can construct just
inside and outside the hemisphere to help you.)
Solution
(b) The wrong way to do this problem is to try to calculate the field everywhere on the xy plane, and then integrate to find the flux. This might sound annoying but doable,
because getting the field at the origin wasn’t so bad. But it gets significantly harder away
from the origin. Fortunately, there is a much easier second method, involving Gauss’s
Law.
We know that ∯𝑆 𝐸⃗ ∙ 𝑑𝑎 =
𝑄𝑒𝑛𝑐
𝜀̥
, so any surface we imagine enclosing our charged
hemisphere will have a flux through it of
𝑄
𝜀̥
. First imagine a surface which just barely
encloses the hemisphere, infinitesimally close to it on both the top and the bottom (and
the rim). It must have a flux through it of
𝑄𝑒𝑛𝑐
𝜀̥
.
How much of this flux goes through the top of the surface, and how much through the
bottom? Let’s compare two little patches of our surface, one just above the hemisphere
and one just below. If the patch is small enough, the hemisphere beneath will “look flat”.
(If not, keep zooming in until it does.) Since it is flat, by symmetry the flux through the
top and bottom must be the same. But this is true for every little patch; there was nothing
special about the one that we chose. That means the flux through a surface just above the
hemisphere must be equal to the flux through a hemisphere just below it.
Now, the flux through the bottom half of the surface is the same as the flux through the xy plane. (Make sure you understand why this is true, it’s important. If it isn’t clear, I’d
come ask about it. Hint: Think about electric field lines. If there are no charges in a
region, these lines aren’t allowed to begin or end.) Hence, the flux through the x-y plane
is
𝑄𝑒𝑛𝑐
2𝜀 ̥
.