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:

The cone has base radius $$R$$ and height $$h$$, and sits in a uniform electric field $$\vec{E}$$ directed parallel to the base. Since there is no enclosed charge, Gauss's law gives zero net flux through the entire closed surface of the cone (curved surface plus the circular base). Therefore the flux entering one side of the cone equals the flux leaving the other side.

To find the entering flux, consider the projection of the cone onto a plane perpendicular to $$\vec{E}$$. The cone's silhouette when viewed along the direction of $$\vec{E}$$ is a triangle with base equal to the diameter $$2R$$ and height equal to $$h$$. The area of this triangle is $$\dfrac{1}{2} \times 2R \times h = Rh$$.

Since the field is uniform, the flux through any closed surface equals the flux through the projected cross-sectional area perpendicular to the field. The entering flux (which equals the leaving flux) is therefore $$\Phi = E \times Rh = EhR$$.