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

Difficult

A

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

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

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

None of the above

Solution

$\vec{E}$ due to infinite plane sheet having surface charge density a on its one face is $\frac{σ}{2 ε_{0}}$
$So {\bar{E}}_{1} = \frac{σ_{1}}{2 ε_{0}} \hat{i} + \frac{σ_{2}}{2 ε_{0}} \hat{j} {\bar{E}}_{2} = - \frac{σ_{1}}{2 ε_{0}} \hat{i} + \frac{σ_{2}}{2 ε_{0}} \hat{j} {\vec{E}}_{3} = - \frac{σ_{1}}{2 ε_{0}} \hat{i} - \frac{σ_{2}}{2 ε_{0}} \hat{j} {\vec{E}}_{4} = \frac{σ_{1}}{2 ε_{0}} \hat{i} - \frac{σ_{2}}{2 ε_{0}} \hat{j}$

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