Current I is flowing through the two materials having electrical … - Vidyadip

MCQ

Current $I$ is flowing through the 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

✓
$I \in_0 (1/ \sigma_2 - 1/\sigma_1)$
B
$\frac{{I{ \in _0}(1/{\sigma _2} - 1/{\sigma _1})}}{4}$
C
$4I \in_0 (1/\sigma_2 - 1/\sigma_1 ) $
D
none of these

✓

Answer

Correct option: A.

$I \in_0 (1/ \sigma_2 - 1/\sigma_1)$

a
$\oint \overrightarrow{\mathrm{E}} \cdot \overrightarrow{\mathrm{dA}}=\frac{q_{\mathrm{in}}}{\varepsilon_{0}}$

$\mathrm{E}_{2} \mathrm{A}-\mathrm{E}_{1} \mathrm{A}=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_{0}}$

$\mathrm{JA}\left(\frac{1}{\sigma_{2}}-\frac{1}{\sigma_{1}}\right)=\frac{\mathrm{q}_{\mathrm{in}}}{\varepsilon_{0}}$

${\varepsilon _0}I\left( {\frac{1}{{{\sigma _2}}} - \frac{1}{{{\sigma _1}}}} \right) = {q_{{\rm{in}}}}$

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