A circular disc of radius 𝑅 carries surface charge density σ(r)=σo(1−rR),where σo is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ϕ0 . Electric flux through another spherical surface of radiusR4 and concentric with the disc is ϕ.Then the ratio ϕ0ϕ is

A circular disc of radius 𝑅 carries surface charge density σ(r)=σ

1.	A circular disc of radius 𝑅 carries surface charge density σ(r)=σo(1−rR),where σo is a constant and r is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is ϕ0 . Electric flux through another spherical surface of radiusR4 and concentric with the disc is ϕ.Then the ratio ϕ0ϕ is
Answer» A circular disc of radius 𝑅 carries surface charge density $σ (r) = σ_{o} (1 - \frac{r}{R})$ ,where $σ_{o}$ is a constant and $r$ is the distance from the center of the disc. Electric flux through a large spherical surface that encloses the charged disc completely is $ϕ_{0}$ . Electric flux through another spherical surface of radius $\frac{R}{4}$ and concentric with the disc is $ϕ$ .Then the ratio $\frac{ϕ_{0}}{ϕ}$ is

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