A parallel-plate capacitor in air has a plate separation of 1.56 cm and a plate
ID: 1588859 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.56 cm and a plate area of 25.0 cm^2. The plates are charged to a potential difference of 200 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an isulator. Determine the charge on the plates before end after immersion. Determine the capacitance and potential difference after immersion. Determine the change in energy of the capacitor. A capacitor of unknown capacitance has been charged to a potential difference of 400 V and then disconnected from the battery. When the charged capacitor is then connected in parallel to an uncharged 10.0-microF capacitor, the potential difference across the combination is 20.0 V. Calculate the unknown capacitance.Explanation / Answer
d = 1.56 cm = 1.56 * 10^-2 m
A = 25.0 cm^2 = 25.0 * 10^-4 m^2
v = 260 V
a)
C = A/d = (8.85*10^-12)*(25*10^-4)/(1.56*10^-2) = 1.42*10^-12 F
Q = CV
Q = (1.42*10^-12 * 260) = 369.2 *10^-12 C
Charge on the Plates before immersion, C = 369.2 pC
Charge will remain same after immersion, as the source is disconnected.
Charge on the Plates after immersion, C = 369.2 pC
b)
C = A/d = 80.4 * (8.85*10^-12)*(25*10^-4)/(1.56*10^-2)
C = 114 * 10^-12 F
Capacitance after immersion, C = 114 * 10^-12 F
V = Q/C
V = (369.2* 10^-12)/(114 * 10^-12)
V = 3.24 Volt
Potential after immersion, V = 3.24 Volt
(c)
Change in the Energy of the Capacitor,
U(i) = (1/2)CV^2 = (1/2)*(1.42*10^-12)(260)^2 = 4.79 * 10^-8 J
U(f) = (1/2)(114.0*10^-12)(3.24)^2 = 5.98 * 10^-10 J
U = 5.98 * 10^-10 - 4.79 * 10^-8 J
U = - 4.73 * 10^-8 J