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A sphere of radius R has charge Q. The electric field strength at distancer>RisEi. What is the ratio Ef/Eiof the final to initial electric field strengths if

(a) Q is halved,

(b) R is halved, and

(c) r is halved (but is still>R)? Each part changes only one quantity; the other quantities have their initial values

Short Answer

Expert verified

(a) The ratio of the final to initial electric field strengths, if Q is halved, is 1/2.

(b) The ratio of the final to initial electric field strengths if radius R is halved is 1.

(c) The ratio of the final to initial electric field strengths, if distance r is halved, is 4.

Step by step solution

01

Given information (part a)

Charge=Qi=QChargeQf=Q2di=r+Rr(Sincer>R)df=r+RrSincer>R

02

Explanation (part a)

sphereofradiusEi=Qi4πε0di2SphereofradiusEf=Qf4πe0df2EfEi=Qf4πe0r2Q4πe0r2EfEi=Q21QEfEi=12

03

Given information (part b)

Sphereofradiusri=RChargeQi=QChargeQf=Qdi=R+rdf=R2+R2+r=R+r

04

Explanation (part b)

sphereofradiusEi=Qi4πε0di2SphereofradiusEf=Qf4πe0df2EfEi=Qf4πε0(R+r)2Qi4πτ0(R+r)2EfEi=Q11QEfEi=11EfEi=1

05

Given information (part c)

Sphereofradiusri=RChargeQi=QSphereofradiusrf=RChargeQf=Qdi=(R+r)rdf=r2+Rr2

06

Explanation (part c)

sphereofradiusEi=Qi4πε0di2SphereofradiusEf=Qf4πe0df2EfEi=Qf4πϵ0r22Qi4πϵ0(r)2EfEi=Q14QEfEi=4

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