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JEE Main 2023Medium

Q.A disc of radius R has surface charge density σ = σ₀ (1 - r/R). The electric field at the centre is

a
σ0/(2ϵ0)\sigma_0 / (2 \epsilon_0)
b
σ0/(4ϵ0)\sigma_0 / (4 \epsilon_0)
c
zero
d
σ0R/(2ϵ0)\sigma_0 R / (2 \epsilon_0)

Correct Answer: Option A

The correct solution involves applying the fundamental concept to derive the final value step by step...

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