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

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EhR

2EhR

4EhR

$\frac{1}{2} EhR$

answer is D.

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Detailed Solution

$ϕ = EA; A \to$ area of the triangle perpendicular to field.

Base = 2R & height = h

$\Rightarrow A = \frac{1}{2} \times 2 R \times h$

$\Rightarrow ϕ = ERh$

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