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Q.A thin rod of length L has linear charge density λ = k x, where x is distance from one end. The electric potential at that end is

a
kL4πϵ0\dfrac{k L}{4\pi \epsilon_0}
b
kL24πϵ0\dfrac{k L^2}{4\pi \epsilon_0}
c
kL2πϵ0ln2\dfrac{k L}{2\pi \epsilon_0} \ln 2
d
kL28πϵ0\dfrac{k L^2}{8\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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