A parallel-plate capacitor in air has a plate separation of 1.41 cm and a plate
ID: 1552558 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.41 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 230 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator
a)Determine the charge on the plates before and after immersion
before=____ pc
after=____pc
b)Determine the capacitance and potential difference after immersion
Cf=____ F
DeltaVf=___V
c)Determine the change in energy of the capacitor
____ nJ
Explanation / Answer
Befor immersion,
C= A/d=1.569 pF
V=230V
Q= CV=360.87pC
After immersion in distilled water, K=80 for water
Distilled water is an insulator. Therefore, after immersion also the charge on the plates remain the same i.e. 360.87pC
b) Dielectric constant of distilled water = 80
Capacitance after immersion = capacitance in air * 80
= 1.569 * 10^-12 * 80
= 125.52pF
Potential difference after immersion = potential difference in air/80
= 230/80 = 2.875volts
(c)Energy stored = Q^2/(2C)
Change in energy stored = (Q^2/2) * (1/C2 - 1/C1)
= (360.87* 10^-12)^2/2 * [1/(125.52 * 10^-12) - 1/(1.569 * 10^-12)]
= - 4.098* 10^-8 J
If only the magnitude of change in energy stored is needed, then
4.098* 10^-8 J