Calculating Electric Field Intensity with Two Spherical Shells

Click For Summary
The discussion focuses on calculating electric field intensity around two spherical shells, with the inner shell carrying a charge of -Q and the outer shell +Q. For regions outside the shells, the electric field is zero, while between the shells, the electric field is influenced by the inner shell's charge. When a point charge is placed at the center, the electric field behaves differently, being non-zero within the inner shell and zero between the shells. It is noted that a point within a spherical shell experiences zero electric field, and the effects of the shells can be treated similarly to point charges if the shells are thin. The analysis applies to both conductors and insulators, assuming uniform charge distribution.
john88
Messages
16
Reaction score
0
Hi

Two spherical shells where the inner has a radius a and the outer a radius of b. The inner has a total charge of -Q whereas the outer shell has a total charge +Q. The question is to calculate the electric field intensity everywhere in space.
My question is now, how do I choose the limits for example a \leq R \leq b I have two examples below

No point charge in the middle

E = 0, 0 \leq R \prec a (not equal to a)

E = -\frac{Q}{4\pi\epsilon_{o}R^3} \bold{R} a \leq R \leq b

E = 0, b \prec R \prec \infty

A point charge in the middle

E = \frac{Q}{4\pi\epsilon_{o}R^3} \bold{R} 0 \prec R \leq a why set equal to a here and not when there aint no point charge in the middle?

E = 0 a \prec R \prec b

E = \frac{Q}{4\pi\epsilon_{o}R^3} \bold{R} b \leq R \prec \infty
 
Physics news on Phys.org
A point within a spherical shell experiences 0N/C. For R between a and b, inner shell is effective and outer shell's net effect is 0N/C...this works for even insulators so long as charge is evenly distributed across the shell (question didn't specify whether it was a conductor or something else). If the thickness of the shells is negligible, then you could consider the effect of the inner shell as that due to a point charge.
 
Last edited:
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
643
Replies
12
Views
2K
  • · Replies 4 ·
Replies
4
Views
969
Replies
4
Views
2K
Replies
12
Views
3K
Replies
4
Views
3K
  • · Replies 17 ·
Replies
17
Views
2K
Replies
6
Views
1K
  • · Replies 16 ·
Replies
16
Views
1K