What is the minimum charge needed for a ball to jump in an electric field?

[Jorgen1224]

I don't really know how to fit what i want to this template, but i'll try.
The thing is that i wonder if anyone can explain to me step-by-step what happened in this solution, because i don't really understand it.

Homework Statement

There are two balls. Upper one(mass m, charge Q) hung on a spring(coefficient k) is being held at the beginning and then it's dropped. Ball on the floor has mass m and charge -Q. Distance between them at the begginning is h.
What min. charge is required so that ball on the floor can jump.

Homework Equations

F_e = k ⋅ q₁ ⋅ q₂ / r²

(There are two different k,one for spring and one from Coulomb's law. Sometimes i don't know which one is which)

The Attempt at a Solution

If i understood correctly l is min. distance between those two balls.

 

[kuruman]

Perhaps it would be best if you explained the solution to yourself with some external guidance.

1. Start with first equation $mg=\frac{Q^2}{4\pi\epsilon_0 l^2}$. What does the left side represent? What about the right side? Are they always equal? What special circumstances must obtain for them to be equal?

2. Look at equation 2. What kind of physical quantities (e.g. speeds, forces, momenta, energies, etc. ) are being added up and set equal to a similar sum of quantities on the other side? Hint: When the sum of all amounts of money that you deposited in your bank account is equal to the sum of all amounts of money that you withdrew in March is equal to your sum of deposits and withdrawals in April, then you say that your money in the bank stays the same (is conserved). What is conserved in this case?

3. Look at equation 3. Where have you seen it before slightly disguised?

Once you answer the above questions, the rest in the solution is algebraic manipulation.

 

[Jorgen1224]

1. I think that it is requirement so that ball can jump ( it has to be ≥ weigh(in this case mg))

2. It is conservation of energy. But why is it k(h - l)²/2 ? Where did that come from? And why in k equation, denominator has changed from h to l?

3. It's the first equation with l on the other side.

Second equation is a mystery to me. Why did someone do it this way? Where did this k(h - l)²/2 come from? Why is it mgl instead of mgh?

 

[kuruman]

You are correct on all accounts. Specifically equation 2 is an energy conservation equation. Remember that energy conservation relates the mechanicl energy at two different points in space A and B, KE_A + PE_A = KE_B + PE_B. What might points A and B be in this case? Hint: There is no kinetic energy at either of these points. The mystery of k(h - l)²/2 will be elucidated if you consider how mechanical energy is transformed from A to B. Obviously, we are talking about transformations of one kind of potential energy to another.

 

[Jorgen1224]

So point A is the beginning and point B when balls are closest they can be to themselves? And how are there different kinds of potential energy?

 

[kuruman]

So point A is the beginning and point B when balls are closest they can be to themselves? And how are there different kinds of potential energy?

Correct. How many kinds of potential energy are there in this problem and what are the expressions for them?

 

[Jorgen1224]

Most certainly simple mgh, but what else? Potential energy of this spring? kx²/2?

 

[kuruman]

Most certainly simple mgh, but what else? Potential energy of this spring? kx²/2?

Yes, potential energy of the spring plus one more. Hint: It's a potential energy that depends on $Q$.

 

[Jorgen1224]

I just looked up such a thing in the internet as i had no idea what it was. As it seems it has to be electric potential energy!

 

[kuruman]

I just looked up such a thing in the internet as i had no idea what it was. As it seems it has to be electric potential energy!

That's what it is. Wasn't it taught to you?

 

[Jorgen1224]

I must've missed this one. Eventually it wasn't introduced to me in school yet, i started attending high school this year. Also i haven't yet read about it in Halliday's Fundamental's of Physics(i bet it's in the 3rd volume).
Anyway, so it is potential energy of spring and mgh changed into electric potential energy and mgl?

 

[Jorgen1224]

Oh i get it. Indeed i now know what this k(h - l)²/2 is.Now if i may, what happened to mgl? Why is it l instead of h? Also in the denominator it is like that too

 

[kuruman]

Follow the algebra.
$mgh-\frac{Q^2}{4\pi\epsilon_0 h}=mgl-\frac{Q^2}{4\pi\epsilon_0 l}+\frac{1}{2}k(h-l)^2$

Using the condition for the ball to jump, replace $\frac{Q^2}{4\pi\epsilon_0}$ with $mgl^2$ in above equation.

$mgh-\frac{mgl^2}{h}=mgl-\frac{mgl^2}{l}+\frac{1}{2}k(h-l)^2$. Drop first two terms on the right side that add to zero.
$mgh-\frac{mgl^2}{h}=\frac{1}{2}k(h-l)^2$. Multiply through by $2h$.
$2mgh^2-2mgl^2=kh(h-l)^2$. Combine terms.
$2mg(h^2-l^2)=kh(h-l)^2$. Factor each side.
$2mg(h-l)(h+l)=kh(h-l)(h-l)$. Divide through by $(h-l)$ which cannot be zero if the ball moves.
$2mg(h+l)=kh(h-l)$.

Solving the last equation for $l$ gives the answer.

 

[Jorgen1224]

2mg(h+l)=kh(h−l).

So from this result, after solving it for l, i get 4th line. Then i take a square root of this which results in 5th line. Why is there a ± sign then?

 

[kuruman]

Then i take a square root of this ...

You take the square root of what? From the condition for the ball to move, you have $Q^2=4 \pi \epsilon_0 l^2$. You take the square root on both sides to get $Q=\sqrt{4 \pi \epsilon_0} l$. You replace $l$ in the expression for $Q$ with the solution for $l$ from $2mg(h+l)=kh(h-l)$. There is a $\pm$ sign in the solution that was provided to you because its author appears to have solved the quadratic formally, thus missing that it can be factored as I showed you in post #13.

 

[Jorgen1224]

Oh wow. I think that now i understand it finally. I actually learned quite a lot from this exercise. You’ve been most helpful, thank you for your time and patience. Not very often i am able to get that much explanation.