A charge Q is uniformly distributed over a long rod AB of length L . The electric potential at the point O lying at a distance L from the end A (where - Physics MCQ JEE Main 2026 | ExamTest.live

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A charge Q is uniformly distributed over a long rod A B of length L . The electric potential at the point O lying at a distance L from the end A (where O, A, B are collinear) is:

Solution & Step-by-step Explanation

1. Let the charge density be λ = \frac{Q}{L} .2. Consider an element d x at distance x from O . The rod extends from x = L to x = 2 L .3. Potential d V = \frac{k d q}{x} = \frac{1}{4 π ϵ _{0}} \frac{λ d x}{x} .4. V = \int_{L}^{2 L} \frac{1}{4 π ϵ _{0}} \frac{Q / L}{x} d x = \frac{Q}{4 π ϵ _{0} L} [ln x]_{L}^{2 L} V = \frac{Q}{4 π ϵ _{0} L} (ln 2 L - ln L) = \frac{Q l n 2}{4 π ϵ _{0} L} .

Try it yourself before checking the explanation above.

A charge Q is uniformly distributed over a long rod A B of length L . The electric potential at the point O lying at a distance L from the end A (where O, A, B are collinear) is:

\frac{Q}{4\pi\epsilon_0 L} \ln 3

\frac{3Q}{4\pi\epsilon_0 L}

\frac{Q \ln 2}{4\pi\epsilon_0 L}

\frac{Q}{8\pi\epsilon_0 L}

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