Electric charges and fields: Charged conducting spheres touching then separated

Click For Summary

Homework Help Overview

The discussion revolves around the behavior of two charged metallic spheres that initially repel each other with a force F. After being touched together, their charges redistribute, leading to a new force of repulsion F'. Participants explore the conditions under which F' may be less than or greater than F, raising questions about charge distribution and the implications for Newton's laws.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss the initial charges of the spheres and how touching them affects their charges and the resulting force of repulsion. There are inquiries about specific charge values and scenarios where the repulsion force could increase after touching.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the mechanics of charge redistribution and its effects on force. Some have provided insights into the mathematical relationships involved, while others express confusion about the implications of these changes.

Contextual Notes

Participants are working under the assumption that both spheres start with positive charges and are questioning how the conservation of charge applies in this context. There is also mention of potential constraints related to homework guidelines and the need for further exploration of the topic.

Ujjwal28
Messages
32
Reaction score
0

Homework Statement


Two charged metallic spheres of same size repel each other by a force F. They are now touched with each other and are then separated to same initial distance. Now the force of repulsion is F'.

Homework Equations


On touching the spheres will reduce their charges... So the new force of repulsion should be less than the original one... Which is the case but there is also one more case in which the new force F' is greater than original force F. How is that possible?

The Attempt at a Solution


Okay so I got the first part correct that F' may be less than F. But I'm stuck here.. How can F' be possibly be larger than F?
When the charges do get distributed on touching.. Any help will be appreciated.
 
Physics news on Phys.org
Ujjwal28 said:

Homework Statement


Two charged metallic spheres of same size repel each other by a force F. They are now touched with each other and are then separated to same initial distance. Now the force of repulsion is F'. [/B]

Homework Equations


On touching the spheres will reduce their charges... So the new force of repulsion should be less than the original one... Which is the case but there is also one more case in which the new force F' is greater than original force F. How is that possible? [/B]

The Attempt at a Solution


Okay so I got the first part correct that F' may be less than F. But I'm stuck here.. How can F' be possibly be larger than F?
When the charges do get distributed on touching.. Any help will be appreciated. [/B]
Hello Ujjwal28. Welcome to PF !

Is this the complete question?

It doesn't seem like they both reduce their charges. If the charges are of the same sign, then the charge one will decrease, and that of the other will increase upon touching.
 
Well yes this is the complete question. And what would he the new charges on a sphere A and B upon touching if.. Initially they carry 2 and 3 coulomb charge respectively? And in what case will the new force of repulsion F' be greater than before?
 
Last edited:
PleAse get back to me
 
Ujjwal28 said:
Well yes this is the complete question. And what would he the new charges on a sphere A and B upon touching if.. Initially they carry 2 and 3 coulomb charge respectively? And in what case will the new force of repulsion F' be greater than before?
What is the result of the touching of two identical spherical conductors, one with initial charge of 2 C the other with 3 C ?
 
Can I get someone to explain to me how this is able to break Newtons action-reaction laws?

How is F able to change just by equaling out the charge? If the balls both start positive then I would think they would have the same force before and after touching.
 
MullaTheMech said:
Can I get someone to explain to me how this is able to break Newtons action-reaction laws?

How is F able to change just by equaling out the charge? If the balls both start positive then I would think they would have the same force before and after touching.
Take some example and calculate the force before and after equalization.

If you still have a question come back, but it's likely that starting a new thread of your own would be more appropriate.
 
Sorry his question intrigued me.
My comments should probably get deleted.
With conservation of charge I thought that two objects with a positive charge would always repel each other with the same force even after touching. I would like to know the equation for equaling out charges.
I'll be waiting for Ujjwal to post what the charge for A & B are after the 2 & 3 coulomb balls touch.
I thought it was going to be simple and you just had to add them up and divide by 2.
 
MullaTheMech said:
Sorry his question intrigued me.
My comments should probably get deleted.
With conservation of charge I thought that two objects with a positive charge would always repel each other with the same force even after touching. I would like to know the equation for equaling out charges.
I'll be waiting for Ujjwal to post what the charge for A & B are after the 2 & 3 coulomb balls touch.
I thought it was going to be simple and you just had to add them up and divide by 2.
Yes, conservation of charge gives the result in your final sentence.
 
  • #10
Right... so I'm thinking I understand this. It seems I've given too much away. I hope you also understand it now Ujjwal.
 
  • #11
Not really! Ok so if two spheres having charges +2 coulomb and +3 coulomb are touched, after touching will the charges become 2.5 coulomb on both? Or am I doing something wrong?
 
  • #12
Ujjwal28 said:
Not really! Ok so if two spheres having charges +2 coulomb and +3 coulomb are touched, after touching will the charges become 2.5 coulomb on both? Or am I doing something wrong?
Yes, that's correct.
 
  • #13
Ok so if that's true then.
Let's take the following case:
1-
Before touching sphere A has a charge of 2C and sphere B has a charge 3C. After touching the charges on both become 2.5C. So..
Repulsion before touching =6.k/r2and repulsion after touching =6.25.k/r2.okay so this proves how F' ( new repulsion) can be greater than F ( original repulsion).But how then F' can be less than F.. Can you please explain me that case?
 
  • #14
Please get back to me
 
  • #15
When would be the repulsion less than before??
 
  • #16
1C×4C=4C
After touching
2.5C×2.5C=6.25C

But I don't think you have to use whole numbers

Is this giving too much away still?
 
  • #17
Well that explains after touching the repulsion increases but there is also a case in which repulsion decreases after touching. Can you explain me that case?
 
  • #18
Whoops I was doing mental math... not good.
I was thinking starting with decimals would get you there. Sorry for confusing you. The guys on this site are usually much better at helping.
 
  • #19
Well that explains after touching the repulsion increases but there is also a case in which repulsion decreases after touching. Can you explain me that case?
 
  • #20
Get back to me please
 
  • #21
Get back to me gentlemen...
 
  • #22
If both charges are positive, Q1*Q2 is never greater than ((Q1+Q2)/2)^2. It is easy to prove. It is also known as "the geometric mean less or equal to the arithmetic mean".
The initial repulsion force is F = k Q1*Q2/r2. After the charges become equal, the force is F ' = k((Q1+Q2)/2)2 ≥ F.
This is valid, when the spheres are so far from each other that the charge distribution is not changed by the other sphere.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
12
Views
2K
Replies
23
Views
5K
  • · Replies 21 ·
Replies
21
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
11
Views
2K
Replies
9
Views
927
  • · Replies 18 ·
Replies
18
Views
3K
  • · Replies 8 ·
Replies
8
Views
3K