How Does Gauss's Law Apply to a Uniformly Charged Sphere?

Click For Summary
SUMMARY

This discussion focuses on applying Gauss's Law to a uniformly charged sphere with a diameter of 10.5 cm and a charge of -22.3 µC. The electric field just inside the paint layer is confirmed to be zero, while the calculations for the electric field just outside the paint layer and 5.00 cm beyond the surface yield a magnitude of approximately -7.27 x 107 V/m. Participants emphasize the importance of precise input and rounding in calculations, particularly when using computational tools.

PREREQUISITES
  • Understanding of Gauss's Law
  • Familiarity with electric field calculations
  • Knowledge of SI units and conversions
  • Proficiency in LaTeX for mathematical expressions
NEXT STEPS
  • Review the derivation of Gauss's Law and its applications
  • Practice electric field calculations for different charge distributions
  • Learn about the significance of precision in numerical computations
  • Explore advanced topics in electrostatics, such as potential energy in electric fields
USEFUL FOR

Students studying electromagnetism, physics educators, and anyone interested in mastering electric field calculations using Gauss's Law.

yaro99
Messages
75
Reaction score
0

Homework Statement


A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a charge of -22.3 µC.

(a) Find the magnitude of the electric field just inside the paint layer.
(b) Find the magnitude of the electric field just outside the paint layer.
(c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Homework Equations


$$E=\frac{q}{4\pi\epsilon_0 r^2}$$


The Attempt at a Solution


I got (a) correct (E=0), but I'm getting wrong answers for (b) and (c)

For (b), the radius is 0.0525m, so:
$$E = \frac{-22.3\times 10^{-6}}{4\pi (0.1025)^2(8.854\times 10^{-12})} = -7.272\times 10^7$$

For (c), r=0.0525+0.05=0.1025, so:
$$E = \frac{-22.3\times 10^{-6}}{4\pi (0.1025)^2(8.854\times 10^{-12})} = -7.272\times 10^7$$

EDIT: for some reason my latex code isn't displaying properly, so here are pics of my equations:
EDIT 2: fixed it
 
Last edited:
Physics news on Phys.org
yaro99 said:

Homework Statement


A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a charge of -22.3 µC.

(a) Find the magnitude of the electric field just inside the paint layer.
(b) Find the magnitude of the electric field just outside the paint layer.
(c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Homework Equations


$$E=\frac{q}{4\pi\epsilon_0 r^2}$$

The Attempt at a Solution


I'm getting wrong answers for (b) and (c)
For (b), the radius is 0.0525m, so:$$E = \frac{-22.3\times 10^{-6}}{4\pi (0.0525)^2(8.854\times 10^{-12})} = -7.272\times 10^7$$
Code:
> (-22.3e-6)/(4*pi*(0.0525^2)*(8.854e-12))
ans =  -7.2717e+07
... hmmm, if this is being computer mediated, then you should check rounding and the exact form of the input, units, stuff like that. Perhaps they'll accept ##\small -7.27\times 10^7##V/m

EDIT: for some reason my latex code isn't displaying properly
That's because you didn't use only latex markup inside the latex tags.
Use the "quote" button (below, right) to see what I did.
 

Similar threads

Replies
6
Views
1K
  • · Replies 12 ·
Replies
12
Views
1K
Replies
10
Views
4K
  • · Replies 11 ·
Replies
11
Views
2K
Replies
23
Views
4K
  • · Replies 26 ·
Replies
26
Views
3K
  • · Replies 6 ·
Replies
6
Views
2K
Replies
4
Views
1K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K