Question

Part A Constants What is the potential difference between the capacitor plates? When an air capacitor with a capacitance of 350 nF (1 nF 10 F is connected to a power supply, the energy stored in the capacitor is 2.05x10-s J. While the capacitor is kept connected to the power supply, a slab of dielectric is inserted that completely fills the space between the plates. This increases the stored energy by 2.48x105 J. You may want to review (Pages 797-803) Submit Request Answer For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Energy storage with and without a dielectric. Part B What is the dielectric constant of the slab? K- nView?assignmentproblemID-107044029#

Part A and Part B please

Answer 1

Air capacitor with a capacitance = Co= 350nF (1 nF= 10^-9F)

The energy stored in the capacitor = Ui =(1/2)CoV^2

Ui= 2.05×10−5J .

(1/2)CoV^2= Ui

V^2= 2Ui/Co

V = sq rt [ 2Ui/Co]

V = sq rt [ 4.1x10^-5/ 350x10^-9 ]

V = 10.82 V

A) The potential difference between the capacitor plates is 10.82 V

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When the capacitor is kept connected to the power supply,the potential difference V remains same and insertion of a slab of dielectric that completely fills the space between the plates increases the capacity

The increased energy = Uf =2.05 ×10−5J + 2.48×10−5 J.

Uf =4.53×10−5 J.

Uf=(1/2)(dielectric const.)CV^2

Uf = Ui*)(dielectric const.)

(dielectric const.) = Uf / Ui

(dielectric const.) =4.53×10−5 /2.05 ×10−5

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(dielectric const.) =2.21

B) The dielectric constant of the slab is 2.21