avolp said:Homework Statement
A parallel plate capacitor with capacitance C is charged to a value q and then isolated. The separation between the plates is then tripled. What was the work required to separate the capacitors?Homework Equations
U=1/2CV^2
Q=CV
C=ε0A/dThe Attempt at a Solution
W=ΔU=1/2C1V1^2-1/2C2V2^2
=1/2C*(Q/C)^2-1/2*(C/3)*(Q/C)^2
=Q^2/3C
I tried this but this wasn't the correct answer and I can't figure out why. Thank you in advance for the help
The formula for calculating the work required to separate parallel plate capacitors is W = 1/2*C*(V2^2-V1^2), where W is the work in Joules, C is the capacitance in Farads, V2 is the final voltage, and V1 is the initial voltage.
The work required to separate parallel plate capacitors is directly proportional to the distance between the plates. This means that as the distance between the plates increases, the work required to separate them also increases.
No, the work required to separate parallel plate capacitors cannot be negative. This is because negative work would mean that energy is being added to the system, which is not possible in this scenario.
If the capacitance of the parallel plate capacitors is increased, the work required to separate them also increases. This is because capacitance is directly proportional to the amount of charge stored on the plates, and more charge requires more work to be separated.
Yes, the work required to separate parallel plate capacitors is affected by the voltage applied. The higher the voltage, the more work is required to separate the plates, as seen in the formula for calculating work.