Find the electrical field of a Spherical Shell

AI Thread Summary
To find the electric field of a thin spherical shell with a radius of 14.0 cm and a total charge of 32.0 µC, the electric field inside the shell at 10.0 cm is zero, as it lies within the charged surface. For a point outside the shell at 20.0 cm, the electric field is calculated using the formula E = (1/4πε₀) * Q/r², resulting in an electric field of approximately 7.192 MN/C. The charge is clarified to be in microcoulombs (µC), confirming the calculation's accuracy. The discussion highlights confusion over the charge unit and emphasizes the need for clarity in such problems. The final answer for the electric field outside the shell is 7.192 MN/C.
RedPhoenix
Messages
26
Reaction score
0

Homework Statement



Consider a thin spherical shell of radius 14.0 cm with a
total charge of 32.0 %C distributed uniformly on its
surface. Find the electric field (a) 10.0 cm and (b) 20.0 cm
from the center of the charge distribution.

Homework Equations



img34.gif


The Attempt at a Solution



Stumped...

I figured that the E(r) was the electrical field of the sphere.

I did 9.9E9(32E-9/14^2) (also did it in meters too)

I guess I am just completely lost in this one.
 

Attachments

  • img34.gif
    img34.gif
    705 bytes · Views: 608
Physics news on Phys.org
Electric field due to a charged sphere at a distance r is given by
E = (1/4πεo)*Q/r^2
In side the charged sphere, the electric field is ...?
 
I got it.

10cm < 14cm, so its inside, meaning = 0

20cm > 14cm, so ke(q/r^2) = 8.99E9(32E-9/.2^2) = 7192, so now I just have to look over where I messed up on the order of magnitude

7.192 MN/C from the center... is the answer!
 
In the problem, the charge is given as 32%C. What is that? micro or nano?
If it is micro, the answer will be correct. You will get 7.192*10^6N/C or 7.192MN/C.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top