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(A) According to Gauss's law, the flux through a closed surface is $\oint \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\varepsilon_0}$.Consider a Gaussian s

Vedclass · Accepted Answer

$I \varepsilon_0 (1/\sigma_2 - 1/\sigma_1)$

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(A) According to Gauss's law, the flux through a closed surface is $\oint \vec{E} \cdot d\vec{A} = \frac{q_{in}}{\varepsilon_0}$.Consider a Gaussian surface in the form of a cylinder enclosing the junction.The electric field in the first material is $E_1 = J/\sigma_1$ and in the second material is $E_2 = J/\sigma_2$, where $J = I/A$ is the current density.The flux through the Gaussian surface is $\Phi = E_2 A - E_1 A = \frac{q_{in}}{\varepsilon_0}$.Substituting the expressions for $E_1$ and $E_2$:$A \left( \frac{J}{\sigma_2} - \frac{J}{\sigma_1} \right) = \frac{q_{in}}{\varepsilon_0}$Since $J = I/A$, we have:$A \cdot \frac{I}{A} \left( \frac{1}{\sigma_2} - \frac{1}{\sigma_1} \right) = \frac{q_{in}}{\varepsilon_0}$$q_{in} = I \varepsilon_0 \left( \frac{1}{\sigma_2} - \frac{1}{\sigma_1} \right)$.

Current $I$ is flowing through two materials having electrical conductivities $\sigma_1$ and $\sigma_2$ respectively $(\sigma_1 > \sigma_2)$ as shown in the figure. The total amount of charge at the junction of the materials is

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