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JEE Advanced 2011Paper 1
Exam
JEE Advanced
Year
2011
Paper
Paper 1
Subject
Physics
Chapter
Electrostatics

JEE Advanced 2011 · Paper 1 · Physics · Electrostatics

Consider an electric field E=E_0 x where E _0 is a constant. The...

Consider an electric field E=E0x^\mathbf{E}=E_0 \hat{\mathbf{x}} where Eˉ0\bar{E}_0 is a constant. The flux through the shaded area ( as shown in the figure) due to this field is Consider an electric field E=E_0 x where E _0 is a constant. The flux through the shaded area ( as shown in the figure)

Options

  1. A2E0a22 E_0 a^2
  2. B2E0a2\sqrt{2} E_0 a^2
  3. CE0a2E_0 a^2
  4. DE0a22\frac{E_0 a^2}{\sqrt{2}}

Correct answer

C. E0a2E_0 a^2

Step-by-step solution

Electric flux, E⋅S\mathbf{E} \cdot \mathbf{S}, or ϕ=EScos⁡θ\phi=E S \cos \theta Here, θ\theta is the angle between E\mathbf{E} and S\mathbf{S}. In this question θ=45∘\theta=45^{{}^{\circ}}, because S\mathbf{S} is perpendicular to the surface. E=E0⇒S=(2a)(a)=2a2∴ϕ=(E0)(2a2)cos⁡45∘=E0a2\begin{gathered}E=E_0 \\ \Rightarrow \quad 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 a² \end{gathered}\thereforeCorrectoptionis(c).AnalysisofQuestion(i)Questionismoderatelytough.(ii)Thegivenshadedareaisarectanglenotasquare.OnesideofthisrectangleisCorrect 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 isaandothersideisand other side is\sqrt{2} a.(iii)Electricfieldisuniform,whosemagnitudeis. (iii) Electric field is uniform, whose magnitude isE_0anddirectionispositiveand direction is positivex.Inuniformelectricfieldwecanuse,. In uniform electric field we can use,\phi=E S \cos \theta
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