Two parallel plane sheets 1 and 2 carry uniform charge densities -1 and -2 as in fig electric field in the region marked II is -1-2

The correct option is D (σ_1 σ_2)/2 ϵ_0E=E_1 E_2 =σ_1/2 ϵ_0 σ_2/2 ϵ_0 E=σ_1 σ_2/2 ϵ_0 ...

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Byju's Answer

Standard XII

Physics

Two Infinite Sheets

Two parallel ...

Question

Two parallel plane sheets 1 and 2 carry uniform charge densities $σ_{1}$ and $σ_{2}$ as in fig electric field in the region marked II is ( $σ_{1}$ > $σ_{2}$ )

- \frac{(σ_{1} + σ_{2})}{2 ϵ_{0}}

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\frac{- (σ_{1} σ_{2})}{2 ϵ_{0}}

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\frac{(σ_{1} + σ_{2})}{2 ϵ_{0}}

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\frac{(σ_{1} - σ_{2})}{2 ϵ_{0}}

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Solution

The correct option is D $\frac{(σ_{1} - σ_{2})}{2 ϵ_{0}}$
$E = E_{1} - E_{2}$
$= \frac{σ_{1}}{2 ϵ_{0}} - \frac{σ_{2}}{2 ϵ_{0}}$
$E = \frac{σ_{1} - σ_{2}}{2 ϵ_{0}}$

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Table-1 Table-2

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Two parallel plane sheets 1 and 2 carry uniform charge densities σ1 and σ2 as in fig electric field in the region marked II is (σ1 > σ2)

The correct option is D (σ1−σ2)2ϵ0E=E1−E2 =σ12ϵ0−σ22ϵ0 E=σ1−σ22ϵ0

Two parallel plane sheets 1 and 2 carry uniform charge densities $σ_{1}$ and $σ_{2}$ as in fig electric field in the region marked II is ( $σ_{1}$ > $σ_{2}$ )

The correct option is D $\frac{(σ_{1} - σ_{2})}{2 ϵ_{0}}$
$E = E_{1} - E_{2}$
$= \frac{σ_{1}}{2 ϵ_{0}} - \frac{σ_{2}}{2 ϵ_{0}}$
$E = \frac{σ_{1} - σ_{2}}{2 ϵ_{0}}$