A one dimensional infinite long wire with uniform linear charge density λ, is placed along the z-axis. The potential difference δV = V(ρ + a) - V(ρ), between two points at radial distances ρ+ a and ρ from the z-axis, where a≪ρ is closest to

Correct option is B

Solution:

Electric Field E⃗\vec{E}E: $$\vec{E} = \frac{\lambda}{2 \pi \epsilon_0 r} \hat{r}$$ 

$$V_2 - V_1 = - \int_{r_1}^{r_2} \vec{E} \cdot d\vec{r}$$ 

$$V_{\rho + a} - V_{\rho} = - \int_{\rho}^{\rho + a} \frac{\lambda}{2 \pi \epsilon_0 r} \, dr = - \frac{\lambda}{2 \pi \epsilon_0} \ln \left( \frac{\rho + a}{\rho} \right) = - \frac{\lambda}{2 \pi \epsilon_0} \ln \left( 1 + \frac{a}{\rho} \right)$$ 

$$\text{Approximation used } \ln(1 + x) \approx x \text{ for small values of } x$$ 

$$V_{\rho + a} - V_{\rho} \approx - \frac{\lambda}{2 \pi \epsilon_0} \frac{a}{\rho}$$​