Effects of Distance and Voltage on Energy Stored in Capacitors

[displayname]

Homework Statement

The distance between two parallel plates is increased while
they remain hooked to a battery. The energy stored in the
capacitor will:

AND

The distance between two parallel plates is doubled while
they remain hooked to a battery. The battery voltage also is
doubled. The energy stored in the capacitor will:

Homework Equations

Q = VC
U₂ = (Q²)/(2C) = 1/2C(V²)

The Attempt at a Solution

As the distance is increased, the capacitance will decrease. Since energy is directly related to capacitance, the energy will also decrease? I'm not sure if I'm right.

 

[gneill]

As the distance is increased, the capacitance will decrease. Since energy is directly related to capacitance, the energy will also decrease? I'm not sure if I'm right.

You are right. Even more right, the energy in a capacitor is (1/2)CV², or (1/2)Q²/C. Make with that what you will!

 

[displayname]

so for the second question, energy increases by a factor of 2. Right?

 

[gneill]

If that's what the equations say, then that's what it is!

 

[2milehi]

Another equation you will need is for capacitance.

where
C is the capacitance;
A is the area of overlap of the two plates;
εr is the relative static permittivity (sometimes called the dielectric constant) of the material between the plates (for a vacuum, εr = 1);
and d is the separation between the plates.

So substitute C with the physical equation into the energy equation

U2 = (Q2)/(2C) = 1/2C(V²)

U2 = 1/2·(εr·A/(4πd))·(V²)

You can draw your own conclusions from here.

Info from Wikipedia -> http://en.wikipedia.org/wiki/Capacitance"