Question

5. From Gauss’s law, the electric field set up by a uniform line of charge is E S 5 a l 2pP0r b r^ where r^ is a unit vector pointing radially away from the line and l is the linear charge density along the line. Derive an expression for the potential difference between r 5 r1 and r 5 r2.

Answer 1

Answer:

 ΔV = 2k λ ln (r₁ / r₂)

Explanation:

The electric potential is

         ΔV = - ∫ E. ds

The expression for the electric field of a charge line is given

           E = 2k λ / r

Let's replace and integrate

         ΔV = - 2k λ ∫ dr / r

         ΔV = -2 k λ ln r

Let's evaluate

       ΔV = - 2k λ ln r₂- ln r₁

       ΔV = -2k λ ln (r₂ / r₁)

       ΔV = 2k λ ln (r₁ / r₂)