figure out C
E = (1/2) C V^2
figure new C and new E
work done = change in energy
A parallel-plate capacitor has plates with an area of 450 cm^2 and an air-filled gap between the plates that is 2.30 mm thick. The capacitor is charged by a battery to 570 V and then is disconnected from the battery.
How much energy is stored in the capacitor? (muJ)
The separation between the plates is now increased to 4.10 mm. How much energy is stored in the capacitor now?(muJ)
How much work is required to increase the separation of the plates from 2.30 mm to 4.10 mm? (muJ)
1 answer
1 answer