Question

a) Two parallel conducting plates are placed next to each other and an electric potential of 20.0 V is applied across the plates. An electron is placed on the plate with the lower potential. The electron is initially at rest. Determine the value of the electric field required to produce an acceleration of -1.6 x10^6 m/s2.

b) Using the conservation of electric energy, determine the value of the final velocity of the electron.

c) Calculate the maximum charge capability of a dielectric-filled capacitor. You can choose any dielectric except polystyrene or air. Each plate has an area of one-quarter of a square meter. The plates of the capacitor are separated by 0.8 mm. The capacitor needs to withstand 20.0 V applied across its plates.

Answer 1

6 ma q = ma ZE - wa 9.1X161.610 E 1.64 10-19 E = 9.1 x 106 N/c. Wosk - K.E 2 DV = ₂ M v² V= 2 qov 2 Cov M 2x1,6 X101920 √7.0341012 Ilx10 31 2.65*106 mis. ( Q=cv Barium titanate = K Abov K= 1250 1250x1x8.85x10 220 4 0.8x10 69 140.62 x 10 69. 14 Me. -3