Capacitor Energy and Charge Relationship

AI Thread Summary
Increasing the charge on a parallel plate capacitor from 3μC to 9μC and the plate separation from 1mm to 3mm affects the energy stored in the capacitor. The relevant equation for energy is U = 1/2 Q^2/C, which requires understanding the relationship between charge, voltage, and capacitance. The capacitance changes with plate separation, and this must be factored into calculations. The discussion highlights confusion regarding how to find capacitance without knowing voltage and the role of permittivity in calculations. Overall, understanding these relationships is crucial for determining the change in stored energy.
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Homework Statement



If you increase the charge on a parallel plate capacitor from 3μC to 9μC and increase the plate separation from 1mm to 3mm, the energy stored in the capacitor changes by a factor of...

Homework Equations



U = 1/2QV
or 1/2CV^2

The Attempt at a Solution



I'm not sure which equation to even use, I'm not to great when it comes to differentiating between electric field and energy of the capacitor.

I think I should probably use 1/2QV but i don't have a voltage to work with so I'm confused.
 
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Use the relation between charge Q, Voltage V, and Capacitance C to eliminate the voltage from the expression for the energy.
 
dauto said:
Use the relation between charge Q, Voltage V, and Capacitance C to eliminate the voltage from the expression for the energy.

so U =1/2 Q^2/C

i'm still not sure how i would find capacitance without using the C=Q/V and not having a voltage?
also does the distance between the plates changing not matter?
 
The capacitance changes if you change the distance, and you should know a formula that connects the two.
 
mfb said:
The capacitance changes if you change the distance, and you should know a formula that connects the two.

oh alright i thought that the permitivity of free space had to be used when a dielectric was present only but that works thanks!
 
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