JEE Advanced · Physics · 17. Electrostatics
Consider an electric field \(\mathbf{E}=E_0 \hat{\mathbf{x}}\) where \(\bar{E}_0\) is a constant. The flux through the shaded area ( as shown in the figure) due to this field is

- A \(2 E_0 a^2\)
- B \(\sqrt{2} E_0 a^2\)
- C \(E_0 a^2\)
- D \(\frac{E_0 a^2}{\sqrt{2}}\)
Answer & Solution
Correct Answer
(C) \(E_0 a^2\)
Step-by-step Solution
Detailed explanation
Electric flux, \(\mathbf{E} \cdot \mathbf{S}\), or \(\phi=E S \cos \theta\) Here, \(\theta\) is the angle between \(\mathbf{E}\) and \(\mathbf{S}\). In this question \(\theta=45^{\circ}\), because \(\mathbf{S}\) is perpendicular to the surface.
\(E=E_0 \)
\( \Rightarrow S=(\sqrt{2 a})(a)=\sqrt{2} a^2 \)
\( \therefore \phi=\left(E_0\right)\left(\sqrt{2} a^2\right) \cos 45^{\circ}=E_0 a^2 \)
\(\therefore\) Correct option is (c).
Analysis of Question
(i) Question is moderately tough.
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is \(a\) and other side is \(\sqrt{2} a\).
(iii) Electric field is uniform, whose magnitude is \(E_0\) and direction is positive \(x\). In uniform electric field we can use, \(\phi=E S \cos \theta\)i
\(E=E_0 \)
\( \Rightarrow S=(\sqrt{2 a})(a)=\sqrt{2} a^2 \)
\( \therefore \phi=\left(E_0\right)\left(\sqrt{2} a^2\right) \cos 45^{\circ}=E_0 a^2 \)
\(\therefore\) Correct option is (c).
Analysis of Question
(i) Question is moderately tough.
(ii) The given shaded area is a rectangle not a square. One side of this rectangle is \(a\) and other side is \(\sqrt{2} a\).
(iii) Electric field is uniform, whose magnitude is \(E_0\) and direction is positive \(x\). In uniform electric field we can use, \(\phi=E S \cos \theta\)i
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