Part A throught C pleaseA parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant K . The magnitude of the charge on each plate is Q . Each plate has area A , and the distance between the plates is d Part A Use Gauss's law to calculate the magnitude of the electric field in the dielectric. Express your answer in terms of the given quantities and appropriate constants. Part B Use the electric field determined in part A to calculate the potential difference between the two plates. Part C Use the result of part B to determine the capacitance of the capacitor. Express your answer in terms of the given quantities and appropriate constants.

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Asked Oct 5, 2019
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Part A throught C please

A parallel-plate capacitor has the volume between its plates filled with plastic with dielectric constant K . The magnitude of the charge on each plate is Q . Each plate has area A , and the distance between the plates is d 

Part A Use Gauss's law to calculate the magnitude of the electric field in the dielectric. Express your answer in terms of the given quantities and appropriate constants. 

Part B Use the electric field determined in part A to calculate the potential difference between the two plates. 

Part C Use the result of part B to determine the capacitance of the capacitor. Express your answer in terms of the given quantities and appropriate constants. 

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Expert Answer

Step 1

Given:

Magnitude of charge on each plate = Q

Area of each plate = A

Dielectric constant = K

Distance between the plate = d

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Let the area of the enclosed charge be A Then genel. = CA' = a Surface charge density A

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Step 2

(a)

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According to Gauss's law: Edd'= Leni Gauss's law in terms of dielectric k: PkE.ds'= Lnct Plugging in the given values in the formula: A' A E dA' Integrating: QA' kE..A'= A& Q E KAE

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Step 3

(b)

The relation between the potential d...

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V = Ed «d = distance between the plate Plugging in the value of E: V =__xd КАt, Od V KAE

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