Uniform charged ball charge density problem

[kirby2]

A 20-cm radius ball is uniformly charged to 80 nC.
(a) What is the ball’s charge density (C/m3)?
(b) How much charge is enclosed by spheres at points 5, 10 and 20 cm
from the center?
(c) What is the electric field strength at points 5, 10 and 20 cm from the
center?

ATTEMPT:

part a: i don't know how to get the answer, but i know A=4(pi)r^2 and i know how to convert the nC to C. but my units of A will be m^2 and i need m^3.

part b: i think the charge at 5, and 10 is zero because all of the charge is on the outside of the ball. so to find the charge at 20cm (surface) i think it would be 100% of the charge so 80nC.

part c: since there's no charge at 5 or 10 cm i think E=0. at 20cm i would use the formula E=kq/r^2.

am i on the right track?

 

[mathman44]

A 20-cm radius ball is uniformly charged to 80 nC.
(a) What is the ball’s charge density (C/m3)?
(b) How much charge is enclosed by spheres at points 5, 10 and 20 cm
from the center?
(c) What is the electric field strength at points 5, 10 and 20 cm from the
center?

ATTEMPT:

part a: i don't know how to get the answer, but i know A=4(pi)r^2 and i know how to convert the nC to C. but my units of A will be m^2 and i need m^3.

part b: i think the charge at 5, and 10 is zero because all of the charge is on the outside of the ball. so to find the charge at 20cm (surface) i think it would be 100% of the charge so 80nC.

part c: since there's no charge at 5 or 10 cm i think E=0. at 20cm i would use the formula E=kq/r^2.

am i on the right track?

The ball is charged everywhere, not just on the surface. If you want charge per unit CUBED, you want to divide the charge by the volume, not the area.

Knowing this, try b) and c) again.

And a hint: since the charge is evenly distributed, to find the charge contained in a PORTION of the sphere, just find the area of that portion, divide by the total volume of the sphere, and that will give you the proportion of the charge contained in that portion.

 

[2milehi]

If the charge is mobile then it will push out to the surface of the sphere.

 

[kirby2]

ok so, retrying part A i used A=(4/3)(pi)r^3 and got A=0.0335. so my new answer is 8E-8/0.0335=2.387E-6. is this right for part A?

 

[bsrishu]

it depends on whether the ball is conducing or non-conducting.
If the ball is conducting then your approach is right. And answer will be in c/m^2.
But as you have mentioned that answer of charge density has been asked in c/m^3 ball is assumed to be non-conducting. Charge densities at inner points will not be zero. You will have to find electric field at inner points by gauss' law

 

[kirby2]

my last response's answer is in C/m^3. (2.387E-6) is this one right?

 

[mathman44]

Seems fine to me.