JEE Main Physics: Flux of electric field Practice Questions

Correct answer: A — Parallel to the surface

Flux is zero when the angle θ between the electric field vector and the area vector is 90° (cos 90° = 0), meaning the field is parallel to the surface. (Note: The correct physical condition for zero flux is when E⃗ ⊥ A⃗, but if E⃗ is parallel to the surface, the flux through that surface is zero). The most general condition for zero flux is E⃗ · A⃗ = 0.

Correct answer: A — E · A · cos 60°

The electric flux is given by Φ_E = E A cos θ, where θ is the angle between E⃗ and A⃗. Given θ= 60°, Φ_E = E A cos 60°.

Correct answer: D — Weber

The SI unit for electric flux (Φ_E) is the Weber (Wb), which is equivalent to N · m²/C².

Correct answer: D — E · A

If E⃗ is perpendicular to S, the angle θ between E⃗ and the area vector A⃗ is 0°, so cos 0° = 1. Thus, Φ_E = E A cos 0° = E A.

Correct answer: B — Electric Flux

Electric flux is the measure of the total electric field lines passing through a given area.

Correct answer: A — By integrating the field magnitude over the area.

When the electric field is non-uniform, the total electric flux through a surface is calculated by integrating the electric field over the surface area: Φ_E = iint_S E · dA.

Correct answer: B — Joule

Electric flux is defined as Φ_E = E⃗ · A⃗. Since E⃗ is in N/C and A⃗ is in m², the unit of flux is N · m²/C or N · m²/C. However, the fundamental SI unit for flux is N · m²/C or Wb (Weber), but often simplified contextually in introductory problems. The most direct dimensional interpretation involving force and area is related to energy concepts, but based on the fields, the dimensional combination is N · m²/C which is equivalent to Wb (Weber). In SI units, N · m²/C is correct, which is related to J/C. For standard JEE definitions involving fields, the unit is Wb (Weber).

Correct answer: C — EA

Since E is perpendicular to A, θ=0°, and cos 0° = 1. Thus, Flux Φ_E = EA.

Correct answer: D — A(E₁ + E₂)/2

When the electric field varies linearly across a surface, the flux is determined by the average field value over the area. Since the field is linear, the average field is (E₁ + E₂)/2. The flux is Φ_E = E₍avg) · A = (E₁ + E₂)/2 A.

Correct answer: C — E A

According to Gauss's Law, the total electric flux (Φ_E) through any closed surface is proportional to the net charge enclosed. If the electric field is uniform and perpendicular to the surface, Φ_E = E A.