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Electric flux passing through a surface of radius r placed in an electric field E with its axis parallel to the field is:

a) E⋅πr²
c) E⋅2πr
c) E⋅2​/πr²
d) E⋅4πr

User Ikbear
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1 Answer

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Final answer:

The electric flux is calculated as the integral of the electric field across a surface, being dependent on the angle between the field and the surface. For a spherical surface with radius r in a uniform electric field, electric flux is not E⋅(4πr²), as it is only true for a sphere encompassing a point charge according to Gauss's law.

Step-by-step explanation:

The electric flux Φ through a surface in an electric field E is given by the integral of E.cos(θ)dA over the surface, where θ is the angle between the field and the normal to the surface. Specifically, for a planar surface perpendicular to a uniform electric field, this simplifies to Φ = EA, where A is the area of the surface. When the surface is parallel to the field, as in the case of the cylindrical sides of a surface, there is no flux because the field lines are not perpendicular to the surface.

In the case of a spherical surface, when dealing with a point charge, the electric field is radial and therefore perpendicular to the surface at every point. This uniformity makes the electric flux through a closed spherical surface independent of its radius. From Gauss's law, the flux through such a sphere is Φ = q₅ₚₙ / ε₀, where q₅ₚₙ is the enclosed charge and ε₀ is the permittivity of free space. Applying this to a sphere of radius r in a uniform electric field, the flux Φ is not simply E multiplied by the sphere's surface area (4πr²), since it only holds when the sphere encloses a point charge.

User Brandon Enriquez
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