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

E_1 nbsp; gt; E_2 [∵ σ_1 nbsp; gt; nbsp;σ_2] ∴ E_net=E_1 E_2 E_net=σ_12ε_0 σ_12ε_0 E_net=(σ_1 σ_2)2ε_0 lt;!–td border: 1px solid #ccc; ...

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Standard XII

Physics

Electric Potential Due to Charged Infinite Sheets

Two parallel ...

Question

Two parallel sheets $1$ and $2$ carry uniform charge densities $σ_{1}$ and $σ_{2}$ $(σ_{1} > σ_{2})$ as shown in figure. The electric field in the region marked $I I$ is

- \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 $\frac{(σ_{1} - σ_{2})}{2 ε_{0}}$

$E_{1} > E_{2} [∵ σ_{1} > σ_{2}]$

$∴ E_{n e t} = E_{1} - E_{2}$

$E_{n e t} = \frac{σ_{1}}{2 ε_{0}} - \frac{σ_{1}}{2 ε_{0}}$

$E_{n e t} = \frac{(σ_{1} - σ_{2})}{2 ε_{0}}$

 Hence, (d) is the correct answer.

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Table-1 Table-2

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

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