Static Charge Induction: Rod and Sphere Experiment

  • Thread starter Thread starter Ammar Qasim
  • Start date Start date
  • Tags Tags
    Charge Static
Click For Summary
The discussion revolves around the principles of static charge induction using a rod and two insulated spheres. When the rod approaches the spheres, positive charges are induced on one sphere while negative charges accumulate on the other, maintaining overall charge conservation. The spheres start neutral, and upon separation, one sphere becomes positively charged while the other remains negatively charged. Participants clarify the movement of charges and correct misunderstandings regarding the final charge states of the spheres. The conversation emphasizes the importance of understanding charge distribution and conservation in electrostatic interactions.
Ammar Qasim
Messages
7
Reaction score
0

Homework Statement



e78-1341-443c-b923-f7d096e7756b&out=http%3A%2F%2Fwww3.0zz0.com%2F2013%2F08%2F13%2F17%2F522810097.png


Homework Equations


The Attempt at a Solution


I think it depends on induction so when the rod is brought close, the spheres will be positively charged. When they move away of it, A will get a negative charge "by induction" and B will get a positive one
 
Physics news on Phys.org
Ammar Qasim said:

Homework Statement



e78-1341-443c-b923-f7d096e7756b&out=http%3A%2F%2Fwww3.0zz0.com%2F2013%2F08%2F13%2F17%2F522810097.png


Homework Equations





The Attempt at a Solution


I think it depends on induction so when the rod is brought close, the spheres will be positively charged. When they move away of it, A will get a negative charge "by induction" and B will get a positive one

Hi Ammar Qasim, Welcome to Physics Forums.

Since the spheres are on insulated stands and the rod never contacts the spheres, charges can only move around, and no new charges will enter the system (conservation of charge). The two spheres start out neutral (zero net charge). Further, charge can only move where there is a conduction path. What does this tell you about the sum of the charges on the two spheres after they are made separate?

When the spheres are in contact and the rod is brought close, make an approximate sketch of the distribution of charges on the spheres. What will it look like?
 
gneill said:
Hi Ammar Qasim, Welcome to Physics Forums.

Since the spheres are on insulated stands and the rod never contacts the spheres, charges can only move around, and no new charges will enter the system (conservation of charge). The two spheres start out neutral (zero net charge). Further, charge can only move where there is a conduction path. What does this tell you about the sum of the charges on the two spheres after they are made separate?

When the spheres are in contact and the rod is brought close, make an approximate sketch of the distribution of charges on the spheres. What will it look like?

Hi gneill, thank you for help.
When rod is brought close the positive charge will move to A and the negative charge will move to B then A will be attracted to the rob holding the positive charge so the answer will be letter B. Did I get it ?
 
Ammar Qasim said:
Hi gneill, thank you for help.
When rod is brought close the positive charge will move to A and the negative charge will move to B then A will be attracted to the rob holding the positive charge so the answer will be letter B. Did I get it ?


No. What happened to the negative charge that you say moved to B? Where does it go if you say that sphere B ends up neutral while sphere A is positive?
 
Last edited:
gneill said:
No. What happened to the negative charge that you say moved to B? Where does it go is you say that sphere B ends up neutral while sphere A is positive?

Oh, you are right. the negative charge will remain at B so the answer will be letter (A)
 
Ammar Qasim said:
Oh, you are right. the negative charge will remain at B so the answer will be letter (A)


Looks good :smile:
 
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
6
Views
2K
Replies
25
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
8K
Replies
8
Views
12K
Replies
9
Views
565
  • · Replies 5 ·
Replies
5
Views
2K