Answer:
[tex]\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}[/tex]
Explanation:
As we know that the charge per unit length of the long cylinder is given as
[tex]\lambda = \frac{q}{L}[/tex]
here we know that the electric field between two cylinders is given by
[tex]E = \frac{2k\lambda}{r}[/tex]
now we know that electric potential and electric field is related to each other as
[tex]\Delta V = - \int E.dr[/tex]
[tex]\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr [/tex]
[tex]\Delta V = -2k \lambda ln(\frac{b}{a})[/tex]
[tex]\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}[/tex]
[tex]\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}[/tex]
