Using the capacitor simulation found here (http://phet.colorado.edu/en/simulatio

Question

Using the capacitor simulation found here (http://phet.colorado.edu/en/simulation/capacitor-lab), I really just need help with understanding what I am observing.

1. With the voltage set at 1.5V, set the plate separation to 5mm and disconnect the battery.

a) What happens to charge when the plate separation is increased? PREDICT why this happens (I saw that the distance between the plates had no effect on the charge when disconnected from the battery. I just cannot understand WHY)

b) What happens to the voltage when the plate separation is increased? PREDICT why this happens. (I'm seeing no change to V, is that correct, and why?)

c) What happens to the strength of the electric field when the plate separation is increased? PREDICT why this happens. (I'm seeing no change to E, is that correct, and why?)

2. Then repeat these steps using the Stored Energy meter

a) What changes do you observe? PREDICT why this happens.

b) Where does this energy go (or come from)?

3. Repeat part 2 above, with the battery connected

a) What changes do you observe? Predict why this happens.

b) Where does this energy go (or come from)?

Explanation / Answer

(1) We know that the Capacitace is defined as
C = eoA/d where A is area and d is distance between plates.
Hence on increasing the distance between the plates the capacitance will decrease.
(a) So when battery is disconnected then the voltage is no more constant. Hence all the change in the capacitance will be take care by the voltage.
Hence the whatever the charge will be stored on the capacitor will remain constant.
(b) We know that
V = Q/C where V is voltage , Q is charge and C is capacitance.
Hence on increasing the distance lead to decrease the capacitance which will increase the voltage.
(C) Electric field remain constant because increasing in the distance and voltage will cancel eachother effect.

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