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Two parallel plate sheets 1 and 2 carry uniform charge densities $σ_{1} a n d σ_{2}$ (see fig.) The electric field in the region marked 1 is ( $σ_{1} > σ_{2}$ )

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$- \frac{σ_{1}}{2 ε_{0}}$

$- \frac{σ_{2}}{2 ε_{0}}$

$\frac{(σ_{1} - σ_{2})}{2 ε_{0}}$

$\frac{(σ_{1} + σ_{2})}{2 ε_{0}}$

answer is D.

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Detailed Solution

Two charges densities is region 1 and 2 is $σ_{1} and σ_{2}$ Where $(σ_{1} > σ_{2})$

Electric potential in region I is

$E_{net} = E_{1} + E_{2} \Rightarrow= \frac{σ_{1}}{2 ε_{0}} + \frac{σ_{2}}{2 ε_{0}} \Rightarrow E = \frac{σ_{1} + σ_{2}}{2 ε_{0}}$

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