oooo TFW F 3:46 PM webassign.net A parallel plate air capacitor has a capacitanc
ID: 1769031 • Letter: O
Question
oooo TFW F 3:46 PM webassign.net A parallel plate air capacitor has a capacitance of 790 pF. The charge on each plate is 2.75 HC. (a) What is the potential difference between the plates? (b) If the charge is kept constant, what will be the potential difference between the plates if the separation is tripled? (c) How much work is required to triple the separation? A capacitor has parallel plates of area 17 separated by 1.6 mm. The space between the plates is filled with polycarbonate (see the table below). (a) Find the permittivity of polycarbonate. c2/(N m2) (b) Find the maximum permissible voltage across the capacitor to avoid dielectric breakdown. (c) When the voltage equals the value found in part (b), find the surface-charge density on each plate and the induced surface charge density on the surface of the dielectric. each plate Home My Assignments ExtensionRequestExplanation / Answer
1)
(a)
V = Q/C = 2.75 * 10^-6/790 * 10^-12 = 0.00277 * 10^6 = 3.48 * 10^3 V
(b)
V = Q/C = Qd/AEo
if seperation is tipled, potential also tripled!
(c)
W = Q(V2 - V1) = QV = 2.75 * 10^-6 * 3.48 * 10^3 = 9.57 * 10^-3 J
(2)
(a)
? = k?o = 2.8*8.85x10^-12 = 24.78 * 10^-12 C^2/N*m^2
(b)
V = Emxd = (3x10^7)(0.0016) = 4.8 *10^4 V
(C)
E = ?/? = V/d =
? = ?V/d = (24.78x10^-12)4.8x10^4/0.0016=7434*10^-7 = 74.34 * 10^-5 C/m^2 ( surface charge density on each plate. )
? - ?' = ?''
?' = ?oV/d = (8.85x10^-12)4.8x10^4/(0.0016) = 26.55x10^-5
? (from part 3) - ?' = ?''
?'' = 74.34 * 10^-5 - 26.55 * 10^-5 = 47.79 *10^-5 C/m^2 ( induced surface-charge density )
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