Conductivities -1 and -2 where -1-2- then total amount of charge at the junction ofmaterials is

The correct option is B positive J = σ E nbsp; J is same in both the regions and σ_1 gt; σ_2. This means E_1 lt; E_2. So, to make E_2 greater than E_1, ...

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Standard XII

Physics

Potential at a General Point Due to a Dipole

conductivitie...

Question

Current I is flowing through two materials having electrical conductivities $σ_{1}$ and $σ_{2}$ where $σ_{1} > σ_{2}$ , then total amount of charge at the junction of materials is

zero

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positive

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negative

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may be positive or negative

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Solution

The correct option is B positive

$J = σ E$
$J$ is same in both the regions and $σ_{1} > σ_{2}$ . This means $E_{1} < E_{2}$ . So, to make $E_{2}$ greater than $E_{1}$ , positive charge accumulates on the boundry of the materials.

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Similar questions

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Table-1 Table-2

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$(B)$ If $σ_{1} + σ_{2} > 0$	$(Q)$ Electric field in region $I$ is zero.
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Current I is flowing through two materials having electrical conductivities σ1 and σ2 where σ1>σ2, then total amount of charge at the junction of materials is

The correct option is B positive J=σE J is same in both the regions and σ1>σ2. This means E1<E2. So, to make E2 greater than E1, positive charge accumulates on the boundry of the materials.

Current I is flowing through two materials having electrical conductivities $σ_{1}$ and $σ_{2}$ where $σ_{1} > σ_{2}$ , then total amount of charge at the junction of materials is

The correct option is B positive

$J = σ E$
$J$ is same in both the regions and $σ_{1} > σ_{2}$ . This means $E_{1} < E_{2}$ . So, to make $E_{2}$ greater than $E_{1}$ , positive charge accumulates on the boundry of the materials.