# [Calc 3] Gauss' Law for Elecricity over a hemisphere

1. Dec 7, 2008

### mjdiaz89

Hello,

Thank you very much for takign the time to help me out. This is an applied calculus 3 (multivariable calc) level problem. I have an answer, but it is nonsensical as the units are not in accordance with units of charge, C.

1. The problem statement, all variables and given/known data
Instructions: Use Gauss' Law for electricity and the relationship $$q=\int\int\int_{Q} \rho dV$$.

Question: For $$E=\left\langle 2xz^{2},2yx^2,2zy^2\right\rangle$$, find the total charge in the hemisphere $$z=\sqrt{R^{2}-x^{2}-y^{2}$$.

2. Relevant equations
$$\nabla \bullet E=\rho/\epsilon_{o}$$

3. The attempt at a solution
Finding the divergence of E:
$$\nabla \bullet E = \frac{\partial 2xz^{2}}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}$$
Subsituting $$z=\sqrt{R^{2}-x^{2}-y^{2}$$ in the partial of x as z is a function of x:
$$= \frac{\partial 2x(R^{2}-x^{2}-y^{2})}{\partial x}+\frac{\partial 2yx^{2}}{\partial y}+\frac{\partial 2zy^{2}}{\partial z}$$
Yeilds:
$$\nabla \bullet E = 2(R^{2}-2x^{2})$$

Now, this means that $$\rho = 2 \epsilon_o (R^{2}-2x^{2})$$

Using the relation $$q=\int\int\int_{Q} \rho dV$$:
and setting up the region of integration:
$$q = 2 \epsilon_o \int^{R}_{-R} \int^{\sqrt{R^{2}-x^{2}}}_{-\sqrt{R^{2}-x^{2}}} \int^{\sqrt{R^{2}-x^{2}-y^{2}}}_{0}(R^{2}-2^{2}) dz dy dx$$

Seeing spherical symmetry, a conversion from rectangular to spherical coordinates will be convenient. *NOTE: My book and teacher use phi to indicate the angle from the z-axis and theta as the angle from the x-axis; like this: http://mathworld.wolfram.com/SphericalCoordinates.html * :
$$R^{2} - x^{2} = \rho^{2} - (\rho sin\phi cos\theta)^{2}$$

Switching limits of integration and substituting $$\rho dV$$

$$q = 2 \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} \int^{R}_{0} \rho^{2} sin\phi (\rho^{2} - (\rho sin\phi cos\theta)^{2}) d\rho d\phi d\theta$$
Distributing and integrating $$d\rhp$$
$$q = 2 \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} [\rho^{5}/5sin\phi - 2\rho^{5}/5sin^{3}\phi cos^{2}\theta]|^{R}_{0} d\phi d\theta$$

After integrating wrt $$d\rho$$ and moving R^5 outside of the iterated integrals, the iterated integral remains as (Notice the weird R^5 units!):
$$q = 2/5 R^{5} \epsilon_o \int^{2\pi}_{0} \int^{\pi/2}_{0} [sin\phi - 2sin^{3}\phi cos^{2}\theta] d\phi d\theta$$

Once all the trigonometric integrations are done (which is just as long as the work up to this point), my finals answer is
$$\frac{4}{15} \pi R^{5} \epsilon_o$$

The units of this answer are $$[R^{5}][\epsilon_o] = m^{5} * \frac{C^{2}}{Nm^{2}}\neq C$$

Im sure there is an error I made as this is a law. However, my teacher was unable to find my error after visiting him in his office. I have been thinking about this problem for a few days now and ...well... Im still without a sensical answer. Any help?

Last edited: Dec 7, 2008
2. Dec 8, 2008

### gabbagabbahey

You should take note here that for E to represent an actual electric field, it must have units of N/C. The only way this can happen is if you assume (sort of hand-wavingly) that there is some constant $$k=1 \frac{\text{N}}{\text{m}^3\text{C}}$$ multiplying this expression. If you were to assume this, then when you found your answer to be proportional to $$\epsilon_0 R^5$$, you would have that it is actually proportional to $$k\epsilon_0 R^5$$ which would give you the correct units of charge. However, since this is a math course and not a physics course I doubt they expect you to correct the units of E in this manner.

There is a mistake on your part though:

When you do this substitution, you are effectively assuming that $$z=\sqrt{R^{2}-x^{2}-y^{2}$$ everywhere in the hemisphere; but that is only true on the surface of the hemisphere. At a general point in the hemisphere, z is independent of x and y, so $$\frac{\partial }{\partial x} (2xz^2)=2z^2$$ and the divergence of E becomes $$2z^2+2x^2+2y^2=2\rho^2$$.

Last edited: Dec 8, 2008