Question

Two infinitely long wires carry linear charge densities +λ and −λ respectively as shown. The potential difference between points A (at a distance a from the first wire) and B (at a distance b from the second wire) is

Answer 1

We know, electric field due to a linear change distribution at any point at a distance r is :

E = \(\frac{λ}{2\pi\epsilon_0r}\)

We know ;

E = -dv/dr

or dv = -Edr

∴ V = -∫Edr

∴ V_A = \(\int\limits_a^{d-a}(\frac{λ}{2\pi\epsilon _0}) \frac{dr}{r}\)

\(=\cfrac{-λ}{2\pi \epsilon _0 }||lnr|^{d-a}_a\)

\(=\frac{λ}{2\pi \epsilon _0}\) {lna - ln (d -a)}

∴ \(V_B \int\limits_{d-b}^b (\frac{-λ}{2\pi \epsilon_0})\frac{dr}{r}\)

\(=\frac{λ}{2\pi \epsilon_0}|lnr|^b_{d-b}\)

\(=\frac{λ}{2\pi \epsilon _0}\){lnb - ln(d-p)}

∴ V_A - V_B = \(\frac{λ}{2\pi \epsilon _0} \,ln(\frac{(d-a)(d-b)}{ab})\)