A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.25 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 260 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.

(b) Determine the capacitance and potential difference after immersion.

(c) Determine the change in energy of the capacitor.
_____nJ

Explanation / Answer

capacitance=electrical permitivity*area/distance between plates

hence before immersion:

capacitance=8.85*10^(-12)*25*10^(-4)/0.0125=1.77 pF

hence charge on the plates of the capacitor before immersion=capacitance*voltage applied

=1.77*260 pC=460.2 pC

as there is no source attached when immersed in water,

charge will remain constant

part a:

before immersion: charge=460.2 pC

after immersion: charge=460.2 pC

part b:

after immersion :

dielectric constant of distilled water=80

hence capacitance after immersion=80*capacitance before immersion

=80*1.77 pF=141.6 pF

charge=460.2 pC

then voltage difference=charge/capacitance

=3.25 volts

part c:
initial energy stored=0.5*capacitance *voltage^2

=0.5*1.77*10^(-12)*260^2=59.826 nJ

final energy stored=0.5*141.6*10^(-12)*3.25^2=0.747825 nJ

then change in energy=final energy-initial energy

=-59.078175 nJ

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