Understanding the Behavior of a Parallel Plate Capacitor: True or False?

[Physics197]

Homework Statement

The diagram shows a parallel plate capacitor with plate area A and spacing d. The capacitor can be connected to a battery of potential difference V by closing the switch. Assume d² << A.

1. The force on plate 1 due to plate 2 is F = Q²/(4πε0d²) î, where Q is the charge on plate
2.If the switch is closed and d is varied, the charge on the plates varies.
3. If the capacitor is initially uncharged and then the switch is closed, a charge −Q = ε0AV/d appears on plate 1.
4. The force between the plates varies as 1/d if the switch is open.
5. If the switch is closed, the electric field at plate 1, due to plate 2, is V/(2d) î

Select true or false.

Homework Equations

The Attempt at a Solution

1. True - from coulombs law (charges will be the same on either plate)
2. i think false, can the charges be changed..
3. true (unless if that plate should be positivly charged)
4. false - coulombs law
5. false, could be true, but can't find an equation that tells me that

 

[Physics197]

http://f.imagehost.org/view/0778/plates

 

[Physics197]

anyone..

 

[collinsmark]

anyone..

You'll probably get better responses if you include all relevant equations in your attempt at a solution. For example, you mentioned "coulomb's law" in a couple of your attempted answers. But Coulomb's law, as it is generally written, applies to point charges. What is the corresponding equation for parallel plates? Things like that might help. Sometimes, explanations may be in order -- for example, which things in a given equation are held constant when the switch is open, and which things are held constant when the switch is closed.