Capacitor behaviour (force, electric field, charge and plate distance)

[bedarder]

Homework Statement

[PLAIN]http://dl.dropbox.com/u/14443433/Screen%20shot%202011-02-25%20at%2011.02.23%20AM.PNG

Homework Equations

Q = C*V
E-field = V/d
F = E*q
C = epsilon-naught*A/d

The Attempt at a Solution

A.
false based on
F = E*q
= Vq/d
= C*V^2/d

B.
true based on the inverse relationship demonstrated above

C.
false because as d is varied, there is a change in force, and this force will affect the charge, q, as per F = Vq/d

D.
false because isn't the electric field constant everywhere, and doesn't depend on distance?

E.
true because
Q = C*V
= epsilon-naught*A*V/d
the negative sign comes into play based on http://www.physics.upenn.edu/courses/gladney/phys151/lectures/images/charging_capacitor_circuit.gif" picture I found on google images. I'm not actually sure why the capacitor charges in that manner, though.

Since B and E were true, my answer was "BE" which is wrong.

Help?

 

[tiny-tim]

hi bedarder!

(have an epsilon: ε and try using the X² and X₂ icons just above the Reply box )

F = E*q
= Vq/d
= C*V^2/d

B.
true based on the inverse relationship demonstrated above

ah, but F = CV²/d has a C in it, which isn't constant

 

[bedarder]

ah, but F = CV²/d has a C in it, which isn't constant

Ok, so using C = ε*A/d, the equation now becomes
F = ε*A*V²/d²
which makes B false.

So then the answer is simply E? I only have one attempt left to get the question right.

 

[tiny-tim]

So then the answer is simply E? I only have one attempt left to get the question right.

i'm accepting no responsibility!

but for what it's worth, your arguments, on a quick perusal, don't have any obvious flaws

 

[bedarder]

i'm accepting no responsibility!

but for what it's worth, your arguments, on a quick perusal, don't have any obvious flaws

I felt bold and decided to go with "E", and got the question right!

Thank you tim!

You may be tiny, but your help was big