Question
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 distance $$L$$ from the end $$A$$ is
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 distance $$L$$ from the end $$A$$ is
A.
$$\frac{Q}{{8\pi {\varepsilon _0}L}}$$
B.
$$\frac{{3Q}}{{4\pi {\varepsilon _0}L}}$$
C.
$$\frac{Q}{{4\pi {\varepsilon _0}L\ln 2}}$$
D.
$$\frac{{Q\ln 2}}{{4\pi {\varepsilon _0}L}}s$$
Answer :
$$\frac{{Q\ln 2}}{{4\pi {\varepsilon _0}L}}s$$
Solution :
Electric potential is given by,
$$V = \int\limits_L^{2L} {\frac{{kdq}}{x} = } \int\limits_L^{2L} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {\frac{q}{L}} \right)dx}}{x} = \frac{q}{{4\pi {\varepsilon _0}L}}\ln \left( 2 \right)} $$
Electric potential is given by,
$$V = \int\limits_L^{2L} {\frac{{kdq}}{x} = } \int\limits_L^{2L} {\frac{1}{{4\pi {\varepsilon _0}}}\frac{{\left( {\frac{q}{L}} \right)dx}}{x} = \frac{q}{{4\pi {\varepsilon _0}L}}\ln \left( 2 \right)} $$
