Homework Help: Calculate the maximum electrical force on a ball

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1. Dec 2, 2016

doktorwho

1. The problem statement, all variables and given/known data
A metal ball of radius $a$ is at a distance $h>>a$ from a very long conducting non-charged horizontal metal surface. Calculate the maximum electrical force on a ball under the condition that $E_{cr}$ the critical electric field of air is not breached.
2. Relevant equations
3. The attempt at a solution

I applied the mirror theorem and placed a ball of charge $-Q$ on the other end, $2h$ distance away from the first.
$F_2=Q_1E_2$ this is the force on ball 1 from ball 2. When $E=E_{crit}$, we have a maximum charge of ball 1 and ball 2 because they are the same. so we would have $F_{max}=Q_{max}E_{crit}$ and since they are the same $Q_{max}=E_{crit}16\pi\epsilon_0h^2$ and when i use that instead of $Q_{max}$ i get that $F_{max}=(E_{crit})^2h^216\pi\epsilon_0$
And somehow my book gives this answer:
$$F_{max}=\pi\epsilon_0a^4(E_{crit})^2/h^2$$ How did they include $a$ and why is it divided by $h^2$ I dont understand their solution, can you make something out of it?

2. Dec 2, 2016

haruspex

The force on the real ball is indeed determined by the field at its location from the other ball. Is that the field that is in danger of producing ionisation there?

3. Dec 2, 2016

doktorwho

I think so, it says to fins the maximum force on the real object so that the critical electrical field is not surpassed. From there i drew the above equations but dont know what this answer is and its given as correct

4. Dec 2, 2016

haruspex

There are two fields acting in the space near the real ball. Ionisation risk is from the total field.

5. Dec 3, 2016

doktorwho

I tried like this:
$E_{crt}=\frac{Q_{max}}{4\pi\epsilon_0a^2}$
$Q_{max}=E_{crt}4\pi\epsilon_0a^2$
$F=\frac{Q^2}{4\pi\epsilon_04h^2}$
$F=\pi\epsilon_oa^4E_{crt}^2/h^2$
By exchanging the Q's with their maximum values. It seems correct, at least coherent with the solution.

6. Dec 3, 2016

haruspex

Right.

Technically, as I mentioned, you should sum the fields from the real and fictitious balls, but the latter will be very small by comparison.