A cone of base radius R and height h is located in a uniform electric field E parallel to its base. The electric flux entering the cone is: (a) 1/2 E h R (b) E h R (c) 2 E h R (d) 4 E h R

Answer: (b). Flux entering = E× (area of the cone projected on a plane normal to E). With E parallel to the base, the projection is a triangle of base 2R and height h: area = 1/2(2R)h = Rh. φin = EhR, so the answer is (b).

A cone of base radius R and height h is located in a uniform electric… | JEE Main 2014 (11 Apr, Shift 1)

Q114JEE Main 2014 · 11 APR · SHIFT 1Medium

Electric Flux & Gauss's Law

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:

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Answer: (b)

Flux entering = $E\times$ (area of the cone projected on a plane normal to $\vec E$ ).
With $\vec E$ parallel to the base, the projection is a triangle of base $2R$ and height $h$ : area $= \frac{1}{2}(2R)h = Rh$ .
$\phi_{in} = EhR$ , so the answer is (b).

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Electric Flux & Gauss's Law question from JEE Main 2014 (11 Apr, Shift 1)

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