​ A circular disc of radius carries surface charge density σ(r)= σ0 (1-r/R), ​

Question

A circular disc of radius R carries surface charge density σ(r)= σ₀ (1-\(\frac{r}{R}\)), where σ₀ 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 radius \(\frac{R}{4}\) and concentric with the disc is  ϕ. Then  the ratio \(\frac{ϕ_0}{ϕ}\) is _________. -

Answer 1

=\(\frac{32}{5}\)

= 6.40