- #1
r16
- 42
- 0
Homework Statement
In the Berkley physics course E&M book (by Purcell) problem 1.16 is giving me some issues.
A sphere of radius a was filled with positive charge at uniform density [tex]\rho[/tex]. Then a smaller sphere of radius a/2 was carved out, as shown in the figure http://marines.case.edu/dods/p116.jpg
What are the direction and magnitude of the electric field at A? at B?
Homework Equations
Gauss's Law
[tex]\int_S \vec{E} \cdot d\vec{A} = 4 \pi q[/tex]
The Attempt at a Solution
I am having difficulty figuring out the strength of the electric field at point A. I know that E inside of a hollow sphere with a constant surface charge [tex]\sigma[/tex]is 0, so I imagined a sphere of radius [tex]a/2[/tex] cut out from the center of a sphere of radius [tex]a[/tex] of a constant charge density. Due to superposition, there is no electric field inside because everything cancels out. Then I imagined moving the hollow sphere up a distance [tex]dr[/tex] along the z axis (the x and y-axis is symmetrical). There is now [tex]2\rho \pi (r (1-b/a))^2 dr[/tex] -where [tex]a[/tex] is the radius of the big sphere and b is the distance the center of the hollow sphere is from the center of the big sphere-difference in charge between the two hemispheres-assuming a right circular cylinder for the differential volume. Now there is not an equal distribution of charge outside the sphere and there should be a resultant electric field inside, pointing in the +z direction because there is more charge in the bottom hemisphere vs the top hemisphere. I have a hard time figuring out how to calculate the magnitude of E at the center of the radius-[tex]a[/tex] sphere over the non-symmetric resultant charge distribution of the sphere.
Per gauss's law, there is no net electric flux through the hollow sphere and thus no charge enclosed (which makes sense). This was my original answer but it doesn't seem right.
For part B, I just assumed that superposition holds even though the geometry of the charge configuration is not completely spherical. The resultant charge [tex]Q = \rho (4/3 \pi r^3 - 4/3 \pi (r/2)^3)[/tex] acts as a point charge and then I applied gauss's law for a point charge. Is that correct as well?
Last edited by a moderator: