thank in advance A parallel-plate capacitor in air has a plate separation of 1.5
ID: 2273611 • Letter: T
Question
thank in advance
A parallel-plate capacitor in air has a plate separation of 1.54 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 255 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine the charge on the plates before and after immersion. before pC after pC Determine the capacitance and potential difference after immersion. Cf = F Delta Vf = V Determine the change in energy of the capacitor. nJExplanation / Answer
A. as we know dat C = Ae/d
= 25*10^-4*8.85*10^-12/1.54*10^-2
= 1.437 * 10^-12 Farad
also Q = CV
Q = 1.437*10^-12 * 255
= 3.664*10^-10 C or 366.4 pC
Since dielectrics water , hence after immersion also , the charge ll remain the SAME
hence
Q after= 366.4 pC
b. K = 80
Capacitance after emission = capacitance * 80
= 1.437 * 10^-12 * 80
= 114.96*10^-12 farad or 115 pf
Potential difference = 255/80 = 3.1875 V
c . Energy = q^2/2C
= {366.4*10^-12}^2 * 0.5(1/115*10^-12 - 1/ 1.437*10^-12)
= - 4.613*10^-8 JOule
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