A parallel-plate capacitor is charged to a potentialdifference of 12 V. The area

Question

A parallel-plate capacitor is charged to a potentialdifference of 12 V. The area on each plate is 20 cm2 andthe spacing between plates is 0.15 mm. Find thefollowing: 1. the electric field between the two plates 2. the total charge on each plate 3. the electric work done on an electron moving from thenegatively charged plate to the positively     charged plate. A parallel-plate capacitor is charged to a potentialdifference of 12 V. The area on each plate is 20 cm2 andthe spacing between plates is 0.15 mm. Find thefollowing: 1. the electric field between the two plates 2. the total charge on each plate 3. the electric work done on an electron moving from thenegatively charged plate to the positively     charged plate.

Explanation / Answer

1.The electric field E = V/d = 8e4(V/m) 2.According to Gauss law E*S=Q/0 so Q=1,42e-9(C) 3.The work is equal to the product of the potential difference andthe magnitude of charge. W=12eV=1,92e-18(J)

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