Surface Volume and Line charge densities, how to solve problems?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
turtieari
Messages
2
Reaction score
0
This question is a perfect example:

A solid sphere 25cm in radius carries 14microC, distributed uniformly throughout it's volume. Find the electric field strength a) 15cm b)25cm and c)50 cm from center.

I know that I need my gaussian surface and I also need
p=q/v where p is the (Qenclosed) in the equation: Qenclosed/epsilon=EA

Could someone help me understand what the charge densities are for. Do I have to subsitute Qenclosed for p or do I just use it as a clue that I need the volume?

Thank you for your time!

Warmest regards,
ARi :"D
 
Physics news on Phys.org
I don't think you actually need to use Gauss's law for this one, TBH. Since it's a sphere with uniform ρ, by the Shell Theorem, it acts like all the charge were centered at, well, the center. I'd solve using

E = q/(4πε₀r^2)

If you really want to use a Gaussian surface, since it's symmetrical:

qenc/ε₀ = EA -> E = 14µC/ε₀(4πR^2)

Where R is .15m+whatever the distance is from the sphere. Since the shell theorem comes from Gauss's Law, sort of, I guess they're actually the same answer, though.

Hope this helps.
 
The shell theorem is new to me. Maybe I don't know with that name. THANK YOU!
 
The shell theorem says if you have a "shell", infinitely thin and spherical with uniform charge density, the charge acts as if all the charge were at the center. It also works with rings, but not in the 3rd dimeension. The corollary is that a solid sphere, being an infinite number of shells, also acts as if it were at the center.

We learned it in AP physics, but I don't know if it's a 'real' theorem.