Physics true/false questions v. isolated parallel plates

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For isolated parallel plates, the capacitance (C) decreases as the distance (d) increases, leading to an increase in stored energy (U). Inserting a dielectric increases capacitance (C) while decreasing the stored energy (U). The charge (Q) on the plates remains constant regardless of changes in distance. The electric field between the plates is uniform and independent of their separation when the plate size is large compared to the distance. The correct statements are A, B, C, and F.
Gonger
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Which of the statements below are true for two oppositely charged, isolated parallel plates? (C is the capacitance, U is the stored energy, +Q and -Q are the charges on the plates.) Note: isolated plates can not lose their charge. (Enter ALL correct statements, e.g., BCD)

A) When the distance is doubled, U increases.
True. I am thinking that if the distance is doubled there is more to store energy in.

B) Inserting a dielectric increases C.
I have no idea on this one.

C) When the distance is halved, Q stays the same.
True. I don't think distance will affect the charge.

D) Increasing the distance increases the electric field.
True.

E) When the distance is doubled, C increases.
False. It doesn't have any affect on the capacitance

F) Inserting a dielectric decreases U.
Again no idea.

G) Inserting a dielectric increases Q.
I just have no clue with dielectrics.

Thanks for any help in advance.
 
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Some forumlas you need to know is

C = \frac {Q}{V} = \epsilon_0 \frac {A}{d}, \ \ U = \frac {Q^2}{2C}

So when d increases, C decreases, then U increase.

When ever you are dealing with a dielectric, the capacitance is given by

C_d = \kappa \epsilon_0 \frac {A}{d}

Where \kappa is a constant called the dielectric constant. This is always greater than 1. This says that C_d > C. Which is exactly why dielectrics are inserted, to increase the capacitance.

That should get you started.
 
cool. I got it thanks.
 
Gonger said:
B) Inserting a dielectric increases C.
True. C \propto \epsilon > \epsilon_0

D) Increasing the distance increases the electric field.
True.
False. The field is uniform between the plates and is independent of the separation where the plate size is large compared to separation. The potential depends upon distance (V = Ed)

E) When the distance is doubled, C increases.
False.
True. C is proportional to d.

F) Inserting a dielectric decreases U.
True. The dielectric reduces the field between the plates. So it reduces the potential (V=Ed),

G) Inserting a dielectric increases Q.
False. Charge is constant.

So: ABCEF

AM
 
Andrew Mason said:
True. C is proportional to d.

I think C \propto \frac {1}{d}.
 
Corneo said:
I think C \propto \frac {1}{d}.
Right you are. So ABCF.

AM
 
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