Why Does the Force Between Two Charged Objects Change After Contact?

In summary, the new force F2 is smaller than F1 because the charge on the two objects has been exchanged.
  • #1
xxchickapooxx
4
0
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I was wondering if you could help me explain a physics problem. The topic is Electrostatics: We have two objects that attract each other and the distance between them is r and the force is F1. Then we put those two objects together and then put them back to the same distance as before (r).

Why is the new force F2 smaller than F1 and why do the two object repel each other now?

Well I think that as you move the objects together, the positively charged particles move to one object and the negatively ones go to the other one. But I do not understand why F2 is smaller than F1? Shouldn't it be the same because you don't change the distance, you don't take any charge away?

Thank youu!
 
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  • #2
Hello chick, :welcome:

Maybe it helps to try a numerical example:

Suppose one is charged +8 and the other (number 2) is charged -4 before contact.
What, do you think, happens when they are brought into contact ?
As you say, charge is exchanged; until ... ? When does the charge stop moving ?
 
  • #3
BvU said:
Hello chick, :welcome:

Maybe it helps to try a numerical example:

Suppose one is charged +8 and the other (number 2) is charged -4 before contact.
What, do you think, happens when they are brought into contact ?
As you say, charge is exchanged; until ... ? When does the charge stop moving ?
Until the charge is the same on both objects?
 
  • #4
Correct (we assume the objects have approximately the same capacity to hold charge).

What are the respective charges then ?
 
  • #5
BvU said:
Correct (we assume the objects have approximately the same capacity to hold charge).

What are the respective charges then ?
Neutral I suppose?
 
  • #6
In post #1 you concluded
xxchickapooxx said:
you don't take any charge away
So if you start with +8 and -4 = +4 in total, and
  1. you don't take any charge away, and on top of that
  2. both objects have equal charge,
then what are the respective charges ?

(and subsequently: what does that mean for the force F2?)

PS have to run now (sports :smile:)
 
  • #7
BvU said:
In post #1 you concluded
So if you start with +8 and -4 = +4 in total, and
  1. you don't take any charge away, and on top of that
  2. both objects have equal charge,
then what are the respective charges ?

(and subsequently: what does that mean for the force F2?)

PS have to run now (sports :smile:)

Are they +6 and -6... Or I think I got really confused
 
  • #8
xxchickapooxx said:
Neutral I suppose?

You start with charges of +8 and -4 before contact. What is the total (that is, NET) charge? What will you see after contact? The answer follows logically from material you have probably already seen (or I hope you have seen).
 
  • #9
xxchickapooxx said:
Neutral I suppose?

Not neutral, but rather equal to each other. (This assumes of course that both objects are conductors having symmetrical geometries. If the objects are of different sizes and shapes it gets more complicated. Let's just assume for this exercise that both objects are identical, conducting spheres). You had it right in post #3.

xxchickapooxx said:
Are they +6 and -6... Or I think I got really confused

Not equal and opposite to each other. Just equal to each other (again this assumes symmetrical geometries and such).

So what's the net charge left over after the objects come in contact with each other? Divide that leftover charge equally between the two. :wink:
 
  • #10
xxchickapooxx said:
But I do not understand why F2 is smaller than F1?

Oh, and by the way, I wanted to point out that in general, [itex] F_2 [/itex] is not always smaller than [itex] F_1 [/itex]. It depends upon the specific initial conditions which force will be larger. Sometimes [itex] F_2 [/itex] will be smaller than [itex] F_1 [/itex], but it can be larger, depending on how the numbers work out.

I hate to bring up new examples before the previous example was finished, but it might help to first consider even simpler examples than the one @BvU introduced. I wholeheartedly apologize now if this ends up adding confusion.

Example 2:
One object has a charge of 8 units and the other has a charge of -8 units. What are the forces before and after the balls come in contact with each other and are separated again? You may answer qualitatively rather than plugging the numbers into formulas.

Example 3:
One object has a charge of 8 units and the other has a charge of -8.001 units. What are the initial and final forces in this case? (Again, qualitative answers are fine.)

Example 4:

One ball initially has a charge of 8 units and the other has a charge of 0 units (i.e., the other ball is initially neutral). What happens this time?
 
Last edited:

Related to Why Does the Force Between Two Charged Objects Change After Contact?

1. What is electrostatics?

Electrostatics is the study of electric charges at rest, and how they interact with each other and with electric fields.

2. How does electrostatics differ from electricity?

Electrostatics deals with stationary electric charges, while electricity deals with the flow of electric charges (i.e. current).

3. What are some real-life applications of electrostatics?

Electrostatics has many practical applications, such as in electrostatic precipitators for air pollution control, inkjet printers, and electrostatic motors.

4. How can I calculate the electric field due to a point charge?

The electric field at a point due to a point charge can be calculated using the equation E = kq/r^2, where k is the Coulomb's constant, q is the charge of the point charge, and r is the distance from the point charge to the point where the electric field is being measured.

5. Can you explain the concept of electric potential?

Electric potential is the amount of work required to move a unit charge from a reference point to a specific point in an electric field. It is measured in volts and can be thought of as the "potential" for a charge to do work. The higher the electric potential, the more work a charge can do. It is a scalar quantity and can be calculated using the equation V = kq/r, where V is the electric potential, k is the Coulomb's constant, q is the charge, and r is the distance from the reference point.

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