# Capacitor Charge and Change in Dielectric

An air-filled capacitor is connected across a 200-V voltage source.

a) If the air-gap between the two plates is 0.5 mm and the area A of the plates is 0.2 m^2, what is the capacitance?

b) After the source fully charges the capacitor, the capacitor is immersed in transformer oil, whose dielectric constant is k = 4.5. What is the change of the electric energy Î”U in the capacitor if the voltage source remained connected during the process?

c) What is the change of the electric energy Î”U in the capacitor if the voltage source disconnected before it is immersed in the oil?

https://brainmass.com/physics/applied-physics/capacitor-charge-change-dielectric-600044

## SOLUTION This solution is **FREE** courtesy of BrainMass!

a. We use the formula to compute the capacitance of a parallel plate capacitor. The dielectric constant of air is approximately equal to1:

Please see attachment for formula and full solution.

b. With dielectric change while the voltage source is remained connected, the voltage through the capacitor is constantly charging the new capacitor, while the capacitance of the capacitor will increase by a factor of k. Therefore, the new energy stored in the capacitor will be:

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