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

DontAngryMe00

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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