A circular disc of radius R carries surface charge density σ ( r ) = σ 0 1 - r R , where σ 0 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 R 4 and concentric with the disc is ϕ . Then the ratio ϕ 0 ϕ is

Step-by-step solution

Consider a ring element of radius r , thickness in disc charge of element d Q = 2 π r d r σ   ⇒ d Q = 2 π r σ d r Total charge, Q = ∫ 2 π r σ 0 1 - r R dr Q = 2 π σ 0 r 2 2 - r 3 3 R ϕ 0 =  charge upto  R ϵ 0 = 2 π σ 0 R 2 2 - R 3 3 R ϕ =  charge upto  R 4 ϵ 0 = 2 π σ 0 R 2 16 × 2 - R 3 64 × 3 R ϕ 0 ϕ = 1 2 - 1 3 1 32 - 1 192 = 1 6 × 32 × 6 5 = 6 . 4