Question 220 of 311

A charge Q is uniformly distributed over a long rod AB of length L as shown in the figure. The electric potential at the point O lying at a distance L from the end A is :

A$\dfrac{\text{Q}}{4\pi {\in }_{0}\text{L}\text{ln}2}$B$\dfrac{\text{Q}\text{ln}2}{4\pi {\in }_{0}\text{L}}$C$\dfrac{\text{Q}}{8\pi {\in }_{0}\text{L}}$D$\dfrac{3\text{Q}}{4\pi {\in }_{0}\text{L}}$

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Answer: (B) $\dfrac{\text{Q}\text{ln}2}{4\pi {\in }_{0}\text{L}}$

$\text{V}=\int _{\text{L}}^{2\text{L}}\dfrac{\text{kdq}}{\text{x}}$

$=\int _{\text{L}}^{2\text{L}}\dfrac{1}{4\pi {\epsilon }_{0}}\dfrac{\left(\dfrac{\text{Q}}{\text{L}}\right)\text{dx}}{\text{x}}$

$=\dfrac{\text{Q}}{4\pi {\epsilon }_{0}\text{L}}\ell \text{n}\left(2\right)$

JEE Main Physics practice question on Electrostatics (Electric Potential and Potential Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.