JEE Mains · Physics · STD 12 - 2. Electric potential and capacitance
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 distance \(L\) from the end \(A\) is
- A \(\frac{{Qln2}}{{4\pi {\varepsilon _0}L}}\)
- B \(\;\frac{Q}{{8\pi {\varepsilon _0}L}}\)
- C \(\;\frac{{3Q}}{{4\pi {\varepsilon _0}L}}\)
- D \(\;\frac{{3Q}}{{4\pi {\varepsilon _0}Lln2}}\)
Answer & Solution
Correct Answer
(A) \(\frac{{Qln2}}{{4\pi {\varepsilon _0}L}}\)
Step-by-step Solution
Detailed explanation
Electric potential is given by, \(V=\int_{L}^{2 L} \frac{k d q}{x}=\int_{L}^{2 L} \frac{1}{4 \pi \varepsilon_{0}} \frac{\left(\frac{q}{L}\right) d x}{x}=\frac{q}{4 \pi \varepsilon_{0} L} \ln (2)\)
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