A parallel-plate capacitor in air has a plate separation of 1.42 cm and a plate
ID: 2113963 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.42 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 270 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume that distilled water is an insulator. (a) What is the charge on the plates before immersion? pC What is the charge on the plates after immersion? pC (b) What is the capacitance after immersion? pF What is the potential difference after immersion? V (c) What is the change in energy of the capacitor? ?U = Uwater - Uair = (nJ)Explanation / Answer
Plate separation d = 1.42 cm = 0.0142 m
Plate area A = 25.0 cm^2 = 0.0025 m^2
Potential difference V = 270 volts
Capacitance C = e0 * A/d = 8.85 * 10^-12 * 0.0025/0.0142
= 1.558 * 10^-12 farad
a)Charge on the plates before immersion = C*V
= 1.558 * 10^-12 * 270
= 420.7 * 10^-12 coulomb
= 420.7 pC
Distilled water is an insulator. Therefore, after immersion also the charge on the plates remain the same i.e. 420.7 pC
b) Dielectric constant of distilled water = 80
Capacitance after immersion = capacitance in air * 80
= 1.558 * 10^-12 * 80
= 124.6 * 10^-12 farad
= 124.6 pf
Potential difference after immersion = potential difference in air/80
= 270/80 = 3.375 volts
c) Energy stored = Q^2/(2C)
Change in energy stored = (Q^2/2) * (1/C2 - 1/C1)
= (420.7 * 10^-12)^2/2 * [1/(124.6 * 10^-12) - 1/(1.558 * 10^-12)]
= - 5.6 * 10^-8 J
If only the magnitude of change in energy stored is needed, then
5.6 * 10^-8 J