2.E Part E The space between the plates is filled with a dielectric which has a

Question

2.E

Part E

The space between the plates is filled with a dielectric which has a dielectric constant of 20, and the new capacitor is charged up to the same voltage as before. How much energy is stored in the capacitor now?

SubmitMy AnswersGive Up

Energy = nJ C che P5-4 Fropar C secure l https master inqphysicscom entProblemID 74171744 PHYS1052 General Physics II PS-S Pre Part A PS-5-Prep SS2013 niticant fl Co Energ My Answers Part C If you plave -3 peC thang the origin, whet is lhe x-uompunerul vi the iu ve on i ium other ihree charges? My Answer -3 that Rin, what the y component vi th he other th n thing of 4 next 0+HC ced on %3D +%2302.

Explanation / Answer

capacitance=epsilon*A/d

where epsilon=electrical permitivity of free space=8.85*10^(-12)

A=area=0.18*0.18=0.0324 m^2

d=distance between plates=1 mm=0.001 m

then capacitance=C=8.85*10^(-12)*0.0324/0.001=2.8674*10^(-10) F

part A:

electric field=voltage/distance

=35/0.001=35000 N/C

part B:

charge on each plate=capacitance*voltage

=2.8674*10^(-10)*35=1.0036*10^(-8) C

part C:

capacitance is 2.8674*10^(-10) F=2.8674*10^(-6) uF

part D:

energy stored=0.5*capacitance*voltage^2

=0.5*2.8674*10^(-10)*35^2=1.7563*10^(-7) J

=175.63 nJ

part E:
new capacitance=dielectric constant*old capacitance

then energy is stored=0.5*new capacitance*voltage^2

=0.5*20*old capacitance*voltage^2

=20*175.63 nJ

=3512.6 nJ

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.