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Q.A rod of length L has linear charge density λ = λ₀ (1 - x/L). The potential at one end is

a
λ0L4πϵ0\dfrac{\lambda_0 L}{4\pi \epsilon_0}
b
λ0L2πϵ0\dfrac{\lambda_0 L}{2\pi \epsilon_0}
c
λ0L4πϵ0ln2\dfrac{\lambda_0 L}{4\pi \epsilon_0} \ln 2
d
λ0L4πϵ0\dfrac{\lambda_0 L}{4\pi \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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