Calculating Forces on Two Charges Above a Conducting Sheet

In summary, the problem involves two charges, +Q and -Q, located a horizontal distance a apart and a vertical distance b above a conducting sheet. The attempt at a solution involves using image charges and an expression for the force, but the modulus and vector components are difficult to determine. The final answer is given as F = kQ^2 times vector ( [1/a^2 - a/(a^2 + 4b^2)^3/2] , 0 , [2b/(a^2 + 4b^2)^3/2 - 1/4b^2] ).
  • #1
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Homework Statement



Two charges +Q and -Q are a horizontal distance a apart and a vertical distance b above a large conducting sheet. Find the components of the forces acting on each charge.

Homework Equations





The Attempt at a Solution



Ok well I've tried to solve this using image charges (i.e. positing 2 imaginary charges below the plate) and have got an expression for F. My problem is that the expression is really horrible so i can't find the modulus of the vector bit and thus write it in components.. have i gone wrong?

my answer is F = kQ^2 times vector ( [1/a^2 - a/(a^2 + 4b^2)^3/2] , 0 , [2b/(a^2 + 4b^2)^3/2 - 1/4b^2] )

Thanks!
 
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  • #2
Those look like vector components from where I'm sitting! :smile:
 

1. How do you calculate the electric force between two charges above a conducting sheet?

The electric force between two charges above a conducting sheet can be calculated by using Coulomb's law, which states that the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

2. What is the significance of the conducting sheet in calculating forces between two charges?

The conducting sheet acts as an equipotential surface, meaning that the electric potential is the same at all points on the sheet. This allows for easier calculation of the electric field and force between the charges, as the presence of the sheet does not affect the electric field lines.

3. How does the distance between the two charges affect the electric force?

The electric force between two charges is inversely proportional to the square of the distance between them. This means that as the distance between the charges increases, the force between them decreases. Conversely, as the distance decreases, the force increases.

4. Can the electric force between two charges above a conducting sheet ever be zero?

Yes, it is possible for the electric force between two charges above a conducting sheet to be zero. This occurs when the charges are placed at equal distances from the sheet and are of equal magnitude, creating a state of equilibrium where the forces cancel each other out.

5. How does the charge of the conducting sheet affect the electric force between the two charges?

The charge of the conducting sheet does not directly affect the electric force between the two charges. However, it does play a role in determining the electric potential at different points on the sheet, which in turn affects the electric field and force between the charges. A higher charge on the sheet would result in a stronger electric field and higher electric force between the charges.

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