# Calculate the maximum electrical force on a ball

• doktorwho
In summary, the maximum electrical force on a ball is determined by the field at its location from the other 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?

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?

haruspex said:
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

doktorwho said:
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.

haruspex said:
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.

doktorwho said:
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.

## 1. How is the maximum electrical force on a ball calculated?

The maximum electrical force on a ball is calculated using the equation F = k*q1*q2/r^2, where F is the force, k is the Coulomb's constant, q1 and q2 are the charges of the two objects, and r is the distance between the two objects.

## 2. What is the Coulomb's constant?

The Coulomb's constant, denoted as k, is a proportionality constant that is used in the calculation of electrical force between two charged objects. Its value is approximately 8.99 x 10^9 N*m^2/C^2.

## 3. How does the distance between the two objects affect the maximum electrical force?

The maximum electrical force between two objects is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases and vice versa.

## 4. What is the unit of measurement for electrical force?

The unit of measurement for electrical force is Newtons (N). This is the same unit used to measure other types of forces.

## 5. Can the maximum electrical force on a ball be negative?

Yes, the maximum electrical force on a ball can be negative. This occurs when the two charged objects have opposite charges and the force between them is attractive instead of repulsive. Negative force values indicate an attractive force, while positive force values indicate a repulsive force.

• Introductory Physics Homework Help
Replies
4
Views
1K
• Introductory Physics Homework Help
Replies
16
Views
276
• Introductory Physics Homework Help
Replies
6
Views
381
• Introductory Physics Homework Help
Replies
4
Views
690
• Introductory Physics Homework Help
Replies
23
Views
1K
• Introductory Physics Homework Help
Replies
43
Views
3K
• Introductory Physics Homework Help
Replies
2
Views
2K
• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
11
Views
914
• Introductory Physics Homework Help
Replies
9
Views
1K