JEE Main 2014 · 11 Apr · Shift 1 · Physics Q16

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PhysicsSingle correct+4 / -1Question 16 of 90

A cone of base radius R and height $h$ is located in a uniform electric field $\vec E$ parallel to its base. The electric flux entering the cone is :

\dfrac12EhR

EhR

2EhR

4EhR

Answer and worked solution

Official answer: B

The flux entering equals $E$ times the cone's cross-section perpendicular to $\vec E$ : a triangle of base $2R$ and height $h$ . So $\Phi=E\cdot\frac12(2R)h=EhR$ .

Topics:Electric flux and Gauss's law

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JEE Main 2014 · 11 Apr · Shift 1, Physics question 16