Please show work, The answers are: a) 0.369 nC b) 0.369 nC c) 0.1184 nF d) Energ
ID: 1430726 • Letter: P
Question
Please show work, The answers are:
a) 0.369 nC
b) 0.369 nC
c) 0.1184 nF
d) Energy (after - before) = -45 nJ
This is just for review and want to know how to do it
3. A parallel-plate capacitor in air has a plate separation of 1.5cm and a plate area of 25 cm2. The plates are charged to a potential difference of 250 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 immersion b) Determine the charge on the plates after immersion c) Determine the capacitance and potential difference after immersion d) Determine the change in energy of the capacitorExplanation / Answer
Plate separation d = 1.50 cm = 0.0150 m
Plate area A = 25.0 cm^2 = 0.0025 m^2
Potential difference V = 250 volts
Capacitance C = 0 A/d = 8.85 * 10^-12 * 0.0025/0.0150
= 1.475 * 10^-12 farad
a)Charge on the plates before immersion = C*V
= 1.475 * 10^-12 * 250
= 369 * 10^-12 coulomb
= 369 pC
Distilled water is an insulator. Therefore, after immersion also the charge on the plates remain the same i.e. 369 pC
b) Dielectric constant of distilled water = 80
Capacitance after immersion = capacitance in air * 80
= 1.475 * 10^-12 * 80
= 118 * 10^-12 farad
= 118 pf
Potential difference after immersion = potential difference in air/80
= 250/80 = 3.125 volts
c) change in energy of capacitor = Uf - Ui
with Uf = (1/2)*Cf*Vf^2
and Ui = (1/2)*Ci*Vi^2
where Ci = initial conductance, which is Q/ Vi
Ci = Q/ Vi = 0.369*10^-9 C / 250V = 1.476*10^-12 F
Vi = 250V
Cf = 118*10^-12 F (from part B)
Vf = 3.125 V (from part B)
plugging in the equation:
Uf - Ui = (1/2)*Cf*Vf^2 - (1/2)*Ci*Vi^2
= (1/2)*( 118*10^-12 F)*( 3.125 V)^2 - (1/2)*( 1.476*10^-12 F)*(250V)^2
= -45 nJ