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(C) Let the linear charge density be $\lambda_0$ at $t=0$ and $\lambda(t)$ at any time $t$.Using Gauss's law,the electric field $E$ at a distance $r$

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(C) Let the linear charge density be $\lambda_0$ at $t=0$ and $\lambda(t)$ at any time $t$.Using Gauss's law,the electric field $E$ at a distance $r$ from the axis is given by $E = \frac{\lambda}{2 \pi \varepsilon r}$.According to Ohm's law,the current density $j$ is $j = \sigma E = \frac{\sigma \lambda}{2 \pi \varepsilon r}$.The rate of change of charge per unit length is equal to the current flowing out through a cylindrical surface of radius $r$ and length $l$:$\frac{dq}{dt} = -j(2 \pi r l) = -\left( \frac{\sigma \lambda}{2 \pi \varepsilon r} \right) (2 \pi r l) = -\frac{\sigma \lambda l}{\varepsilon}$.Since $q = \lambda l$,we have $\frac{d\lambda}{dt} = -\frac{\sigma \lambda}{\varepsilon}$.Solving this differential equation,we get $\lambda(t) = \lambda_0 e^{-\sigma t / \varepsilon}$.Substituting this into the expression for current density,we get $j(t) = \frac{\sigma \lambda_0}{2 \pi \varepsilon r} e^{-\sigma t / \varepsilon} = j_0 e^{-\sigma t / \varepsilon}$.This represents an exponentially decaying function,which corresponds to the graph shown in option $C$.

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