A parallel plate capacitor in air has a plate separation d = 1.5 cm and a plate

Question

A parallel plate capacitor in air has a plate separation d = 1.5 cm and a plate area A = -25.0cm2. The plates are charged to a potential difference V = 400.0 V and then disconnected from the battery. The capacitor is then immersed in distilled water (k = 80.0). Assume the distilled water is an insulator. Charge Q on the capacitor before immersion. Qbefore Charge Q on the capacitor after immersion. Calculate capacitance C after immersion. Calculate potential difference V after immersion. The change in the energy stored in the capacitor due to immersion.

Explanation / Answer

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complete solution is...

capacitance before immersion = (epsilon0*A)/d = 1.475E-12 F

C)after immersion Cnew = kC0 = 80*1.475E-12 =1.18E-10 F
D) Volatge difference = Q/Cnew = 400/80 = 5 V
E) initial energy = 0.5*(1.475*10^(-12))*400*400 = 1.18e-7
final energy = 0.5*(1.18*10^(-10))*5*5 = 1.475e-9

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