Electric Flux Through a Cone or Disc: Concepts, Formulas & Problems

Electric Flux Through A Cone Or Disc is a crucial concept in JEE Main Physics, measuring how much electric field “passes through” surfaces like discs and cones. It is vital for understanding symmetry and applying Gauss’s law to non-flat geometries. Both open and curved surfaces feature heavily in JEE numerical problems and require a stepwise approach to visualize the setup and use the right formula.

Electric flux, symbolized by Φ, quantifies the “flow” of electric field E through a given area. The SI unit is volt–metre (V·m) or newton–metre² per coulomb (N·m²/C). For planar surfaces, Φ = E·A·cosθ, but for a disc or cone, careful attention to the surface orientation and field direction is needed.

Concept and Calculation of Electric Flux

Electric flux through a surface is defined as the dot product of the electric field vector and the area vector, or the surface integral for curved surfaces. For any surface, Φ = ∫_S E · dA. For a uniform field crossing a flat area A at an angle θ, it simplifies to Φ = E·A·cosθ. The area vector is always perpendicular to the surface.

This definition applies to open surfaces like discs and cones, as well as closed surfaces like spheres and cubes. In JEE Main, recognizing whether a surface is open or closed is important because Gauss’s law strictly applies to closed surfaces only, but flux can be calculated for any defined surface in space.

Electric Flux Through A Disc: Formula and Examples

To calculate electric flux through a disc in a uniform field, use Φ = E·A·cosθ, where E is the field magnitude, A = πR² is the disc area, and θ is the angle between the field and the disc’s normal. If the field is perpendicular, cosθ = 1, so Φ = EπR². If parallel, flux is zero.

For a point charge q located on the disc’s axis at distance a above the centre, the total flux through the disc is given by:

These results help quickly identify if flux is zero (field parallel), maximum, or partial based on angle and charge position. The distinction between disc and ring surfaces is essential, as a ring (being narrower) "cuts" less field lines.

For more guidance on area vectors and flux sign convention, see Electric Flux and Area Vector.

Electric Flux Through a Cone or Conical Surface

The electric flux through a cone involves the symmetrical arrangement of field lines and the use of solid angles. For a uniform field E making angle α with the cone’s axis, the curved surface area A = πrl (where r is base radius, l is slant height). The area vector is perpendicular to the curved surface at each point. By symmetry, the total flux through the lateral surface is:

Φ = E × curved area × cos(θ), where θ is the angle between E and the local normal. For a cone with apex at the origin and field along the axis, the total flux through the curved part is often given by Φ = EA·cosα (if the whole surface is considered).

• For a cone in a uniform E-field along axis: Φ = E × πrl
• If E makes angle α with axis: Φ = E × πrl × cosα
• If apex charge q, flux through cone subtending solid angle Ω: Φ = (q/4πε₀)·(Ω/4π)

Be careful: when the cone is open and field lines do not enter through the base, only the curved surface flux is counted. Always check the symmetry and whether the region is part of a full closed surface, which would let you use Gauss’s law for easier calculation.

Contrast this with Moment of Inertia of a Cone which focuses on mass distribution, not electric field lines.

Flux Through Various Surfaces: Key Contrasts

JEE often asks how electric flux differs for discs, cones, rings, cylinders, and closed surfaces. The main differences are whether the surface is open or closed, the alignment of E, and if any charges are enclosed.

For cylinders, the flux through curved or flat surfaces differs depending on field orientation. Check Electric Flux Through Cylinder for full derivations.

Worked Example: JEE Flux Problem

Example: A disc of radius 0.2 m is placed in a uniform electric field of 100 N/C. The field makes a 30° angle with the disc’s normal. Calculate the flux through the disc.

• Area, A = πR² = π × (0.2)² = π × 0.04 ≈ 0.126 m²
• Φ = E·A·cosθ = 100 × 0.126 × cos(30°) ≈ 100 × 0.126 × 0.866 ≈ 10.9 V·m
• Use SI (V·m) and always check the angle!

See more sample numericals in Gauss Law for practice with symmetry and different surface choices.

Common Pitfalls and Applications

Students often confuse when flux is zero: it happens when the electric field is parallel to an open surface (cosθ = 0), like a disc lying flat in an E-field along its own plane. Another trap is confusing open versus closed surfaces for applying Gauss’s law. Remember, only total flux through a closed surface depends solely on the enclosed charge.

• Always use the correct area vector direction for open surfaces.
• For cones and discs, flux does not depend on “enclosed charge” unless surface is closed.
• For ring versus disc, a ring in uniform E-field along axis has zero flux.

Applications include determining charge in a region, checking field uniformity, and visualizing field penetration in capacitor plates or detectors. For a full catalogue of surface types and formulae, see Electric Flux Through a Cube and the comprehensive Physics Formulas page.

Mastering the calculation of electric flux through a cone or disc equips JEE aspirants to handle symmetric and non-symmetric surfaces, link theory to problem solving, and spot zero, maximum, or partial flux cases. For more on core electrostatics concepts, trust the expert theory and examples on Vedantu.