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(C) Consider a Gaussian surface of radius $r$ $(a < r < b)$ and length $L$ between the two cylinders.From Gauss's Law for a line charge density $\lamb

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$\frac{2 \pi \sigma}{\ln (b / a)} V$

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(C) Consider a Gaussian surface of radius $r$ $(a From Gauss's Law for a line charge density $\lambda$,the electric field $E$ at distance $r$ is $E = \frac{\lambda}{2 \pi \varepsilon_0 r}$.The potential difference $V$ is given by $V = \int_{a}^{b} E \, dr = \int_{a}^{b} \frac{\lambda}{2 \pi \varepsilon_0 r} \, dr = \frac{\lambda}{2 \pi \varepsilon_0} \ln(\frac{b}{a})$.Thus,$\lambda = \frac{2 \pi \varepsilon_0 V}{\ln(b/a)}$.The current $I$ flowing through a cylindrical shell of radius $r$ and length $L$ is $I = J \cdot A = (\sigma E) \cdot (2 \pi r L) = \sigma (\frac{\lambda}{2 \pi \varepsilon_0 r}) (2 \pi r L) = \frac{\sigma \lambda L}{\varepsilon_0}$.The current per unit length is $i = \frac{I}{L} = \frac{\sigma \lambda}{\varepsilon_0}$.Substituting the value of $\lambda$: $i = \frac{\sigma}{\varepsilon_0} \cdot \frac{2 \pi \varepsilon_0 V}{\ln(b/a)} = \frac{2 \pi \sigma V}{\ln(b/a)}$.

Two long coaxial and conducting cylinders of radius $a$ and $b$ are separated by a material of conductivity $\sigma$ and a constant potential difference $V$ is maintained between them by a battery. Then the current per unit length of the cylinder flowing from one cylinder to the other is -

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