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PhysicsElectrostaticsGauss LawHard2 minPYQ_2020
PhysicsHardnumerical

A circular disc of radiusRcarries surface charge densityσ(r)=σ01-rR, whereσ0is a constant andris the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely isϕ0. Electric flux through another spherical surface of radiusR4and concentric with the disc isϕ. Then the ratioϕ0ϕis

Answer:
6.40
Solution:

Consider a ring element of radius r, thickness in disc charge of element dQ=2πrdrσ dQ=2πrσdr

Total charge, Q=2πrσ01-rRdr

Q=2πσ0r22-r33R

ϕ0= charge upto Rϵ0=2πσ0R22-R33R

ϕ= charge upto R4ϵ0=2πσ0R216×2-R364×3R

ϕ0ϕ=12-13132-1192=16×32×65=6.4

Stream:JEE_ADVSubject:PhysicsTopic:ElectrostaticsSubtopic:Gauss Law
2mℹ️ Source: PYQ_2020

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