Electric field from concentric spheres

Click For Summary
SUMMARY

The electric field between two concentric spherical shells with radii R1 and R2, each carrying a total charge Q, can be determined using Gauss' law. The electric field inside the shells is zero, while the field outside can be expressed as E = Q / (4πr²ε₀), where r is the distance from the center. If the charge distribution is not uniform, no general formula applies. The discussion emphasizes the importance of mathematical expression in physics problems.

PREREQUISITES
  • Understanding of Gauss' law
  • Familiarity with electric fields and charge distributions
  • Knowledge of vector mathematics in physics
  • Concept of surface charge density
NEXT STEPS
  • Study Gauss' law in detail
  • Explore electric field calculations for non-uniform charge distributions
  • Learn about the implications of spherical symmetry in electrostatics
  • Investigate the relationship between gravitational and electric fields in spherical shells
USEFUL FOR

Physics students, electrical engineers, and anyone studying electrostatics or preparing for exams in electromagnetism.

tony873004
Science Advisor
Gold Member
Messages
1,753
Reaction score
143
Two concentric spherical surfaces with radii R1 , R2 each carry a total charge Q. What is the electric field between the two shells?

I don't know what kind of answer they are expecting. Do I just describe it? Here's my attempt:

The field lines from the inner shell will point away from the inner shell towards the outer shell. The field lines from the outer shell will point away from the outer shell towards the inner shell. Since the outer shell has the same total charge as the inner shell, but it is spread out more over the larger surface area, the point where the field lines meet will be closer to the outer shell.

\frac{{4\pi r_1^2 }}{{4\pi r_2^2 }} = \frac{{r_1^2 }}{{r_2^2 }}
So the field lines from the inner sphere will be \frac{{r_1^2 }}{{r_2^2 }}
stronger than from the outer sphere. So the distance is <br /> \frac{{\frac{{r_1^2 }}{{r_2^2 }}}}{{1 + \left( {\frac{{r_1^2 }}{{r_2^2 }}} \right)}}<br /> times the distance between the spheres, closer to the outer sphere.

Is this right? Is this even the way I should express the answer?

 
Physics news on Phys.org
tony873004 said:
don't know what kind of answer they are expecting. Do I just describe it? Here's my attempt:
The field or the electrical intensity is a mathematical quantity, a vector in fact, and should be written as such.

The field lines from the inner shell will point away from the inner shell towards the outer shell. The field lines from the outer shell will point away from the outer shell towards the inner shell. Since the outer shell has the same total charge as the inner shell, but it is spread out more over the larger surface area, the point where the field lines meet will be closer to the outer shell.
The description is wrong.

Is this right? Is this even the way I should express the answer?
Not right, and no, you should express your answer mathematically. Read up on Gauss' law and fields due to charged shells.
 
Thanks, Shooting star.

Reading up on it, it seems that there are no field lines inside a shell. So would that mean that the outer shell can simply be ignored? Is the answer simply <br /> \overrightarrow E = \frac{Q}{{4\pi r^2 \varepsilon _0 }}{\rm{\hat r}}<br /> ?
 
tony873004 said:
Two concentric spherical surfaces with radii R1 , R2 each carry a total charge Q. What is the electric field between the two shells?

This was the original framing of the problem. Here, we have tacitly assumed (and the problem-maker probably implied) that the surface charge density on the shells are uniform. In that case, the expression you have given is correct.

On the other hand, in the problem it is only mentioned that the the total charge is Q on each surface. If it is not uniformly distributed on each surface, then there is no general formula to describe the field.
 
It wouldn't be the first time this book expected me to assume something. This problem reminds me of the gravity analog. If all Earth's mass were concentrated in a shell with Earth's diameter, gravity would be 1g on the outside surface and 0 everywhere inside.

Thanks for your help.
 

Similar threads

  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 23 ·
Replies
23
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
Replies
6
Views
1K
  • · Replies 17 ·
Replies
17
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
12
Views
2K