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 :

Electric potential $$dV$$ due to a small charge element $$dq$$ at a distance $$r$$ is $$dV = \frac{1}{4\pi\epsilon_0} \cdot \frac{dq}{r}$$

Consider a small element of length $$dx$$ on the rod at a distance $$x$$ from the point $$O$$. The charge on this small element is $$dq = \lambda dx = \frac{Q}{L} dx$$

The electric potential $$dV$$ at point $$O$$ due to this element is $$dV = \frac{1}{4\pi\epsilon_0} \cdot \frac{\frac{Q}{L} dx}{x}$$

To find the total potential $$V$$, we integrate from the nearest end ($$A$$) to the farthest end ($$B$$). Distance of $$A$$ from $$O = L$$ and Distance of $$B$$ from $$O = L + L = 2L$$

$$V = \int_{L}^{2L} \frac{Q}{4\pi\epsilon_0 L} \cdot \frac{dx}{x}$$

$$V = \frac{Q}{4\pi\epsilon_0 L} \ln \left( \frac{2L}{L} \right) = \frac{Q \ln 2}{4\pi\epsilon_0 L}$$