The electric potential due to an infinite uniformly charged line with linear charge density λ at a distance r, with the reference point at r₀, is given by:
a) (2λ) / (4πϵ₀ r)
b) (2λ) / (4πϵ₀ r₀)
c) (2λ [ln r₀ - ln r]) / (4πϵ₀)
d) (2λ ln r) / (4πϵ₀ r)

The electric potential VV at a distance rr from an infinitely long uniformly charged wire with linear charge density λ\lambda is given by the expression:
V(r)=λ2πϵ0ln⁡(rr0)V(r) = \frac{\lambda}{2\pi \epsilon_0} \ln \left( \frac{r}{r_0} \right)
where:
• λ\lambda is the linear charge density of the wire,
• ϵ0\epsilon_0 is the permittivity of free space,
• rr is the distance from the wire where the potential is being calculated,
• r0r_0 is the reference distance where the potential is chosen to be zero.
Explanation:
The electric potential due to a uniformly charged infinite wire is derived from the electric field due to the wire, which is radially symmetric around it. The electric field at a distance rr from the wire is E=λ2πϵ0rE = \frac{\lambda}{2\pi \epsilon_0 r}, and the potential is the integral of the electric field.
The expression for the potential is:
V(r)=∫r0rE dr=∫r0rλ2πϵ0r drV(r) = \int_{r_0}^{r} E \, dr = \int_{r_0}^{r} \frac{\lambda}{2\pi \epsilon_0 r} \, dr
This gives:
V(r)=λ2πϵ0ln⁡(rr0)V(r) = \frac{\lambda}{2\pi \epsilon_0} \ln \left( \frac{r}{r_0} \right)
Correct Answer:
c) 2λ[ln⁡r0−ln⁡r]4πϵ0\frac{2 \lambda \left[ \ln r_0 - \ln r \right]}{4 \pi \epsilon_0}