physics exam 2

Topics: Energy, Capacitor, Work Pages: 2 (279 words) Published: November 21, 2014
An ideal air-filled parallel-plate capacitor has round plates and carries a fixed amount of equal but opposite charge on its plates. All the geometric parameters of the capacitor (plate diameter and plate separation) are now DOUBLED. If the original energy stored in the capacitor was U0, how much energy does it now store?

C= Aε/d
wher C= capacitance, A= area, d= distance between plates
let original radius be r
so original area A= πr2
noe r is doubled and d is also doubled
so new area A2=π(2r)2= 4A
also new distance d2= 2d
so new capacitance Cn=4A/ε(2d) = 2A/εd = 2C
as only geometric changes are done but charges on plates will rtemain same so original energy stored in capacitor Uo= 0.5(Q2/C)
new energy stored in capacitor Uf = 0.5(Q2/Cn) = 0.5[0.5(Q2/C)] = 0.5(Uo)= Uo/2 so new energy stored in capacitor Uf =Uo/2
final answer
a charge of 2.00 uC flows into the plates of a capacitor when it is connected to a 12.0 V battery. How much work was done in charging this capacitor?

Work done = 1/2*Q*V =1.2*10^-5 Joule.

A cylindrical capacitor is made of two thin-walled concentric cylinders. The inner cylinder has radius r1 = 4.0 mm, and the outer one a radius r2= 8.0 mm. The common length of the cylinders is L = 150 m. What is the potential energy stored in this capacitor when a potential difference 4.0 V is applied between the inner and outer cylinder? (k = 1/4??0 = 8.99 × 109 N ? m2/C2)

2,236 answers
2,236 answers Top Subjects: Precalculus, Physics, Chemistry
C = 2πεL/ln(r2/r1) (ε= 8.85x10^-12F/m)
= [2π(8.85x10^-12)*150]/ln(0.008/0.004)
=  [2π(8.85x10^-12)*150]/ln2
= 1.2x10^-8 F
PE = 0.5CV^2
= 0.5*[1.2x10^-8]*4^2
= 9.6x10^-8 J
Energy density (u) = (€oE^2)/2,
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