Calculate the maximum electrical force on a ball

[doktorwho]

Homework Statement

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.

Homework Equations

3. The Attempt at a Solution [/B]
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 don't understand their solution, can you make something out of it?

 

[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?

 

[doktorwho]

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?

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 don't know what this answer is and its given as correct

 

[haruspex]

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 don't know what this answer is and its given as correct

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

 

[doktorwho]

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

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.

 

[haruspex]

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.

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.