Download Spherical charge distribution 2013

Spherical charge distribution 2013
The problem:
Find the electric field due to a sphere that has a charge density of
= where a is a constant.
Technical notes:
1) Must break up into two calculations, one involving the field outside of the sphere and the other
involving the field inside the sphere.
2) Since we’re considering a sphere, spherical coordinates are the best choice for doing this
problem.
3) Will have to perform a triple integral to determine the total charge of the sphere
Physics notes:
1) Will need to use the shell theorem. I won’t prove it here, but Wikipedia’s article should satisfy
any curiosity (http://en.wikipedia.org/wiki/Shell_theorem). Note that this article considers the
gravitational force but the same idea holds for the electric force (and thus the electric field).
Paraphrased, the shell theorem states:
a) Outside a spherically symmetric charge distribution, the distribution can be treated as a
point particle of total charge located at the center.
b) Inside a shell, no net force is exerted on any object inside the shell.
2) When considering the electric field inside the sphere, the charge density decreases with
increasing radius
3) The electric field is continuous at the boundary (although there will be a cusp).
Solution
1. Find an expression for a volume element in spherical coordinate.
2. Use the volume element and the given charge density to calculate the total charge of the sphere
(triple integral).
3. Apply the Shell theorem (part a) to treat the sphere as a point particle located at the origin &
find the electric field due to this point particle.
4. Repeat steps 1 & 2 to find an expression for the charge within a radius r < R (i.e. inside the
sphere).
5. Treat the enclosed charge as a point particle located at the origin (Shell theorem part a) and
calculate the electric field as a function of r. The remaining charge outside the radius r is
spherically symmetric shell and therefore can be ignored (Shell theorem part b).
6. Sanity check
7. Plot the electric field
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Spherical charge distribution 2013
Step 1: The volume element in spherical coordinates
When we’re in Cartesian coordinates, a volume element is straightforward (probably since this is the
coordinate system that we’ve used for most of our lives). However, for curved objects, sometimes
Cartesian system is ill-adept for calculations. Below is a diagram that tries to present how to determine
a volume element in spherical coordinates by first finding a small area (dA) on the surface of a sphere
(Figure 1).
Figure 1: Volume element in spherical coordinates
The volume element in spherical coordinates can then be expressed as:
=
sin(θ)drdθdφ
Please note that there is a confusing use of the two angles θ & φ; in physics θ is the angle between the
vector and the z-axis whereas φ is the angle between the vector and the x-axis in the x-y plane (Figure
2A); in mathematics these two angles are swapped (Figure 2B).
Figure 2: Spherical coordinates in physics and math
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Spherical charge distribution 2013
Step 2: The integral (r>R)
=
Plugging in the expression for the volume element:
=
The standard practice is to evaluate the innermost integral (i.e. dφ). Fortunately none of the variables,
(or limits of integration) depend on φ, so the integration is fairly straightforward.
( | )
=
The next integral (dθ) is slightly more difficult due to the ‘sinθ’ term.
=2
(−
=2
)|
Lastly we have the (dr) integration to perform. Remember that the charge density is a function of ‘r’!
=4
=4
=2
|
=2
Now we can use the idea of the shell theorem to model the spherical charge as a point charge at the
origin and the total charge of the sphere to calculate the outside electric field.
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Spherical charge distribution 2013
Step 3: Electric field outside the sphere
⃗=
⃗=
̂
(2
) ̂
Step 4: Now we attack the charge inside the sphere (r<R). It’s the same process as Step 2 but note that
the limit of integration for dr is not the radius of the sphere! Note that I do not need to consider the
charge present at a radius > r (shell theorem part b).
=
( | )
=
=2
(−
=2
)|
=4
=4
=2
|
=2
Step 5: Find the electric field due to the charge contained within a radius r
⃗=
⃗=
̂
(2
⃗=2
) ̂
̂
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Spherical charge distribution 2013
So inside the sphere the electric field is constant!
Step 6: One good sanity check is to make sure that the answers to Steps 3 & 5 agree when you set r = R.
⃗
|
(2
)|
= ⃗
̂ =2
(2
2
̂
)=2
=2
Good, they match!
Step 7: Plotting the electric field (for completeness).
Prepared by Greg Robison
Phys 241 Graduate TA
August 31, 2013
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