IIT-JEE 2011 · Paper 1 · Physics Q26

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PhysicsSingle correct+3 / -1Question 26 of 69

Consider an electric field $\vec E=E_0\hat x$ , where $E_0$ is a constant. The flux through the shaded area (as shown in the figure) due to this field is

A shaded square-sided parallelogram with corners (0,0,0), (0,a,0), (a,a,a) and (a,0,a), tilted at 45° between the x-y and y-z planes.

2E_0a^2

\sqrt2E_0a^2

E_0a^2

\dfrac{E_0a^2}{\sqrt2}

Answer and worked solution

Official answer: C

Area vector. The parallelogram is spanned by $(0,a,0)$ and $(a,0,a)$ :
$\vec A=(0,a,0)\times(a,0,a)=(a^2,0,-a^2).$

Flux. $\vec E\cdot\vec A=E_0a^2$ .

(C)

Topics:Electric flux and Gauss's law

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IIT-JEE 2011 · Paper 1, Physics question 26