Minimum force to separate two oppositely-charged ions

In summary, the conversation is discussing the solution to a physics problem involving finding the minimum force required to pull two oppositely-charged ions apart. The attempted solution involved differentiating the potential energy and setting it equal to zero, but there was uncertainty about where the solution went wrong. Suggestions were made to plot the graph and differentiate V'(r) to obtain V''(r) and set it equal to zero, but it was pointed out that this would only find the peak attractive force and not the minimum force required. Another approach was suggested, involving finding the critical force that will carry the particle to infinity without residual kinetic energy, but it was noted that this would lead to a complex equation.
  • #1
Shomari
3
0

Homework Statement


Part (d) is the bit I am having trouble doing (the question is in the attachment).

Homework Equations


Apart from what is shown, I also have to use F = dV(r) / dr

The Attempt at a Solution


For Part (b), I showed this by differentiating the potential energy and equating this differential to zero. From there I rearranged the equation and solved for r.
For Part (c), I had a = r and formed the equation Vmax(r) = A*((1/m) - 1), and stated this as being the energy required to pull the two oppositely-charged ions apart.
For Part (d), I stated that the Force applied = Vmax(r) / Distance of separation, in order to find the Distance of separation I equated the potential energy to zero (where there would be zero potential energy between the two ions) and tried to solve for d. However, in doing so I got:
d = exp{[(m-1)/(m+1)]*ln(a/m)}
as the Distance of separation. I feel like this is the wrong way to answering the question, but I am unsure where I went wrong.
 

Attachments

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  • #2
Shomari said:
For Part (d), I stated that the Force applied = Vmax(r) / Distance of separation.
Why would it be that? What is the relationship between force and potential at a point?
 
  • #3
haruspex said:
Why would it be that? What is the relationship between force and potential at a point?
I thought if Work Done = Force * Distance then I could rearrange that equation to get Force = Work Done / Distance. And have (I probably shouldn't call it Vmax(r)), Vmin(r) = Work Done
But the Force equals -dV(r) / dr, at a point. I could try differentiate my value of Vmin(r) with respect to r, but my value for Vmin(r) equals A*((1/m) - 1), and differentiating that should give me 0.
 
  • #4
Shomari said:
Work Done = Force * Distance
That is only true if the force is of constant magnitude and always acts in the same direction as the element of displacement. More generally it is ##W=\int\vec F.\vec {ds}##.
 
  • #5
The problem statement does ask for the constant force needed. The problem with the OP's approach is that the distance is infinite.

@Shomari, try choosing a value for ##m## and plotting V'(r) to see how the function behaves. That might give you an idea of how to solve the problem in general.
 
  • #6
vela said:
The problem statement does ask for the constant force needed.
Good point - I was not allowing for gained KE helping the particle get through the peak opposing force.
 
  • #7
vela said:
The problem statement does ask for the constant force needed. The problem with the OP's approach is that the distance is infinite.

@Shomari, try choosing a value for ##m## and plotting V'(r) to see how the function behaves. That might give you an idea of how to solve the problem in general.
Okay, thanks for that idea. I plotted the graph and put in values for A, a and m. I didn't read the question properly, but I understand that if I want to know the minimum force required then I need to differentiate V'(r) to obtain V''(r), and set this equal to 0 and solve for r.
Once I solve for r I then substitute that value of r into the equation for V'(r).
However, I've gone wrong somewhere whilst differentiating. If ##V'(r) = A(\frac{-a} {r^2} + \frac{(\frac{a} {r})^m} {r})##, then
##V''(r) = A(\frac{-m(\frac{a} {r})^m - (\frac{a} {r})^m} {r^2} + \frac{2a} {r^3})##.
And since ##V''(r) = 0##, ##(\frac{-m(\frac{a} {r})^m - (\frac{a} {r})^m} {r^2} + \frac{2a} {r^3}) = 0##, but when I rearrange to solve for r I obtain
##ln(r) = ln(a) + \frac{ln(m+1)} {m+1} - \frac{ln(2)} {m+1}##.

I don't think applying exponentials to both sides to obtain r, and substituting this into V'(r) will lead me to the correct solution, but I'm unsure where I went wrong in my working out. I attached a picture of my working out as well.
 

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  • #8
Shomari said:
I understand that if I want to know the minimum force required then I need to differentiate V'(r) to obtain V''(r), and set this equal to 0
Doesn't that find the peak attractive force? As @vela pointed out to me, this is not what the question asks for. If a constant force is applied, and the attractive force increases initially (as it does in your plot) then the particle may accelerate at first. It thus may acquire enough KE to carry it through a range in which the attractive force exceeds the constant force.

Suppose a constant force F=F(r) is just sufficient to move it from a to r. So at r there is no residual KE. Thus (r-a)F=V(r)-V(a). Can you see how to find the critical F that will carry it to infinity (with or without residual KE there)? Unfortunately this leads to an awful equation. It is equivalent to finding the tangent from the equilibrium point to the V(r) curve for r>a.

That said, it seems that the given answer is the peak attractive force, not the minimum constant force!
 
Last edited:
  • #9
Shomari said:
when I rearrange to solve for r I obtain
Try that step again. You should get an m-1 exponent.
 

1. What is the minimum force required to separate two oppositely-charged ions?

The minimum force required to separate two oppositely-charged ions depends on the magnitude of the charges and the distance between the ions. The greater the charges and the closer the ions are to each other, the stronger the force needed to overcome their attraction and separate them.

2. How is the minimum force to separate two oppositely-charged ions calculated?

The minimum force to separate two oppositely-charged ions can be calculated using Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. Can the minimum force to separate two oppositely-charged ions be measured?

Yes, the minimum force to separate two oppositely-charged ions can be measured using specialized equipment such as an ion trap or an atomic force microscope. These instruments can apply precise forces to ions and measure the amount of force required to separate them.

4. How does the minimum force to separate two oppositely-charged ions relate to their stability?

The minimum force to separate two oppositely-charged ions is directly related to their stability. If the force applied to the ions is greater than the minimum force required to separate them, the ions will be pulled apart and become unstable. This is why ions with stronger charges or those that are closer together are more stable.

5. What factors can affect the minimum force to separate two oppositely-charged ions?

The minimum force to separate two oppositely-charged ions can be affected by several factors, including the magnitude of the charges, the distance between the ions, and the presence of any surrounding molecules or ions that may interfere with the attractive force between them. Temperature and the type of medium in which the ions are located can also affect the minimum force required to separate them.

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