A plastic disk of radius R = 64.0 cm is charged on one side with a u

allhvasstH

allhvasstH

Answered question

2020-11-11

A plastic disk of radius R = 64.0 cm is charged on one side with a uniform surface charge density σ=6.28 fCm2, and then three quadrants of the disk are removed. The remaining quadrant is shown in Figure. With V = 0 at infinity, what is the potential dueto the remaining quadrant at point P, which is on thecentral axis of the original disk at a distance D = 25.9cm from the original center?
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Answer & Explanation

Gennenzip

Gennenzip

Skilled2020-11-12Added 96 answers

Potential energy at P overall
V=(σ2σ0)(σR2+D2)Dσ 
=6.28fCm2 
=6.281015 Cm2σ0 
=8.851012C2Nm2 
R=64cm=0.64m 
D=25.9cm=0.259m 
PLUG THE VALUES WE GET V 
After getting V POTENTIAL DUE TO ONE QUARTER =V4 
=1.357105

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