Q.A thin rod of length L has linear charge density λ = λ₀ (1 + x/L). The electric field at a point on the axis at distance a from the centre is

\dfrac{\lambda_0}{4\pi \epsilon_0} \left( \dfrac{1}{a - L/2} - \dfrac{1}{a + L/2} \right)

\dfrac{\lambda_0 L}{4\pi \epsilon_0 a (a + L)}

\dfrac{\lambda_0}{4\pi \epsilon_0 a}

\dfrac{\lambda_0 (2a + L)}{4\pi \epsilon_0 a^2}

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

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