\(V =\int_{ a }^{ b } E \cdot d \ell=\frac{\lambda}{2 \pi \varepsilon_0 r } \ln \left(\frac{ b }{ a }\right)\)

Now, I \(=\sigma \int \overrightarrow{ E } \cdot \overrightarrow{ dA }=\sigma \int \frac{\lambda}{2 \pi \varepsilon_0 r } \cdot 2 \pi dr =\frac{\sigma \lambda}{\varepsilon_0}\)

From \((1)\): \(I =\frac{2 \sigma \pi \varepsilon_0}{\varepsilon_0 \ln ( b / a )}=\frac{2 \pi \sigma}{\ln ( b / a )} v\)