A parallel-plate capacitor in air has a plate separation of 1.41 cm and a plate

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 240 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 ?Vf ___= V (c) Determine the change in energy of the capacitor. ____ nJ

Explanation / Answer

Befor immersion,
C= A/d=1.569 pF

V=240V

Q= CV=376.6pC

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. 376.6 pC

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
= 240/80 = 3 volts

(c)Energy stored = Q^2/(2C)
Change in energy stored = (Q^2/2) * (1/C2 - 1/C1)
= (376.6 * 10^-12)^2/2 * [1/(125.52 * 10^-12) - 1/(1.569 * 10^-12)]
= - 4.463 * 10^-8 J
If only the magnitude of change in energy stored is needed, then
4.463 * 10^-8 J

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