Effect of squeezing a capacitor on energy stored

[aarond90]

2 parallel capacitors, 3uF each, connected in series to 10V battery. One capacitor's separation is then reduced to 30% of its initial value.
What is the total energy stored of the capacitors?

C tot = 1 / ((1/3e-6) + (1/3e-6)) = 1.5e-6F

U initial = (0.5)(1.5e-6)(100) = 7.5e-5 J

After squeeze:

C(1 final) = 3e-6/0.3 = 1e-5 F

C tot (final) = 1 / ((1/1e-5) + 1/3e-6)) = 2.307 e-6 F

U final = 0.5(2.307e-6)(100) = 1.154e-4 J

Delta U = U final - U initial = 4.038e-5 J

- Is this correct?

Many thanks

 

[anathema]

for your post! I can confirm that your calculations and approach are correct. The total energy stored in the capacitors will increase after the separation of one capacitor is reduced, as the overall capacitance of the circuit increases. This means that the capacitors can store more charge at the same voltage, resulting in an increase in energy stored. Keep up the good work!