Minimum force to separate two oppositely-charged ions

[Shomari]

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 V_max(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 = V_max(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.

 

[haruspex]

For Part (d), I stated that the Force applied = V_max(r) / Distance of separation.

Why would it be that? What is the relationship between force and potential at a point?

 

[Shomari]

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 V_max(r)), V_min(r) = Work Done
But the Force equals -dV(r) / dr, at a point. I could try differentiate my value of V_min(r) with respect to r, but my value for V_min(r) equals A*((1/m) - 1), and differentiating that should give me 0.

 

[haruspex]

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 $W=\int\vec F.\vec {ds}$.

 

[vela]

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.

 

[haruspex]

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.

 

[Shomari]

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.

 

[haruspex]

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!

 

[haruspex]

when I rearrange to solve for r I obtain

Try that step again. You should get an m-1 exponent.