A parallel-plate capacitor in air has a plate separation of 1.77 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.77 cm and a plate area of 25.0 cm^2. The plates are charged to a potential difference of 245 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 after pC pC (b) Determine the capacitance and potential difference after immersion. Cf = F delta Vf = F V (c) Determine the change in energy of the capacitor. nJ

Explanation / Answer

a)

Capacitance without dielectric

Cbefore=eoA/d =(8.85*10^-12)*(25*10^-4)/0.0177

Cbefore=1.25 pF

Charge

Qbefore=CbeforeV=245*1.25

Qbefore=306.25 pC

after immersion also charge remains constant

Qafter=306.25 uC

b)

Capacitance with dielectric

Cf=KCbefore=80*1.25 *10-12

Cf=1*10-10 F

Potential difference after immersion

Vf=Q/Cafter=306.25/100

Vf=3.0625 Volts

c)

Initial energy stored

Ui=(1/2)CV2=(1/2)(1.25*10-12)(245)2=3.75*10-8 J

Final energy stored

Uf=(1/2)(3.0625)2(1*10-10)=4.69*10-10 J

so change in energy

dU=Uf-Ui=(4.69*10^-10)-(3.75*10^-8)

dU=-37.03 nJ

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