A capacitor has a plate area of 0.0075 m2 and a free charge of magnitude q on ea

Question

A capacitor has a plate area of 0.0075 m2 and a free charge of magnitude q on each plate. The space between plates has been filled with a dielectric that has a bound surface charge of magnitude 2q/3 on either side. When the dielectric material is removed, the electrical breakdown of air is observed between the plates.

Part A: What is the smallest possible value of q if the plates are isolated?

Part B: What is the smallest possible value of q if the plates are connected to a battery?

Problem 26.26 A capacitor has a plate area of 0.0075 m and a free charge of magnitude q on each plate. The space between plates has been filled with a dielectric that has a bound surface charge of magnitude 2q/3 on either side. When the dielectric material is removed, the electrical breakdown of air is observed between the plates. Part A What is the smallest possible value of q if the plates are isolated? Express your answer to two significant digits and include the appropriate units Armin =) Value Units Submit My Answers Give Up Part B What is the smallest possible value of q if the plates are connected to a battery? Express your answer to two significant digits and include the appropriate units Imin qmin= Value Submit My Answers Give Up

Explanation / Answer

part a:

let dielectric constant be k.

For a parallel plate capacitor with dielectric present,

bound surface charge=(dielectric constant-1)*free surface charge/(dielectric constant)

2q/3 =(k-1)*q/k

solving for k, k = 3

dielectric strength of air=3*10^6 V/m

Electric field E=surface charge density/epsilon

E =charge/(area*epsilon) = Q/eoA

E = q/(0.0075*8.85*10^(-12))

Dielectric field strength of air = 3*10^6

q=0.0075*8.85*10^(-12)*3*10^6

q = 2*10^-7 C (ANSWER)

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part b:

if a battery of emf V is connected to the plates,

electric field=V/d

even after removal of the dielectric ,

this value of electric field will remain present in between the plates.

so for dielectric break down,

E = V/d=3*10^6...(1)

q = k * Eo

q=0.0075*8.85*10^(-12)*3*10^6*3

q = 5.97 *10^-7 C (ANSWER)

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