A parallel plate capacitor consists of two plates each of area 12.4 cm2.. The di

Question

A parallel plate capacitor consists of two plates each of area 12.4 cm2.. The distance between the plates is 3.0 mm. There is a vacuum between the capacitor plates. The voltage difference between the plates is (triangle symbol)V=1200 V. A very small sphere with charge q=+1.20 uC and a mass of2.48 x 10-5 kg is placed inside the capacitor next to the positive plate. The sphere is then released from rest. What is the change in potential energy of the charge as it moves from the positive to the negative plate of the capacitor? The answer is either: A. -1.20ueV B. +1.20 ueV C. -1.44mJ D. +1.44mJ E. -1.20keV F. +1.20keV G. -1.00GJ H. +1.00GJ

Explanation / Answer

E = V / d = 1200 / 3x10^-3 = 36x10^5 N/C

Change in potential energy =work done

= F*X = EQ*X

= 36x105x1.20x10-6x3x10-3   12.96 mj

answer A

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.