.oooo TFVV 11:43 PM 61 %. D webassign.net A parallel-plate air capacitor has a c
ID: 1769022 • Letter: #
Question
.oooo TFVV 11:43 PM 61 %. D webassign.net A parallel-plate air capacitor has a capacitance of 790 pFThe charge on each plate is 2.75 C. (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 cm2 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 (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. Home My Assignments Extension RequestExplanation / 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 )
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.