Two infinite sheets having charge densities $σ_{1}$ and $σ_{2}$ are placed in two perpendicular planes whose two-dimensional view is shown in Fig. The charges are distributed uniformly on the sheets in electrostatic equilibrium condition. Four points are marked I, II, III, and IV. The electric field intensities at these points are ${\vec{E}}_{1}, {\vec{E}}_{2}, {\vec{E}}_{3} a n d {\vec{E}}_{4}$ respectively. The correct expression for the electric field intensities is

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$| {\vec{E}}_{1} | = | {\vec{E}}_{2} | = \frac{\sqrt{σ_{1}^{2} + σ_{2}^{2}}}{2 ε_{0}} \neq | {\vec{E}}_{4} |$

$| {\vec{E}}_{2} | = | {\vec{E}}_{4} | = \frac{\sqrt{σ_{1}^{2} + σ_{2}^{2}}}{2 ε_{0}}$

none of these

$| {\vec{E}}_{1} | = | {\vec{E}}_{2} | = | {\vec{E}}_{3} | = | {\vec{E}}_{4} | = \frac{\sqrt{σ_{1}^{2} + σ_{2}^{2}}}{2 ε_{0}}$

answer is C.

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Detailed Solution

Using the principle of superposition E due to infinite plane sheet having surface charge density con its one face is $\frac{σ}{2 ε_{0}}$ .

${\vec{E}}_{1} = \frac{σ_{1}}{2 ε_{0}} i + \frac{σ_{2}}{2 ε_{0}} j$ , ${\vec{E}}_{2} = - \frac{σ_{1}}{2 ε_{0}} i + \frac{σ_{2}}{2 ε_{0}} j$ , ${\vec{E}}_{3} = - \frac{σ_{1}}{2 ε_{0}} i - \frac{σ_{2}}{2 ε_{0}} j$ , ${\vec{E}}_{4} = \frac{σ_{1}}{2 ε_{0}} i - \frac{σ_{2}}{2 ε_{0}} j$