Q4. Assertion: the electric flux through a small area can be negative. Reason: flux is E ΔS cos θ, and cos θ is negative when θ is more than 90°. Are both true, and does the reason explain the assertion?
Both are true, and the reason explains the assertion: on a closed surface, for example, lines entering it give negative flux (NCERT Eq. (1.11), page 22).
Q5. A circular loop of radius 0.1 m lies in a uniform field of 2 × 103 N C−1, with its normal at 60° to the field. Find the flux through it.
ΔS = π(0.1)2 ≈ 0.0314 m2, so ϕ = (2 × 103)(0.0314)(0.5) ≈ 31 N m2 C−1 (NCERT Eq. (1.11), page 22).
Q6. Two ways of reading E ΔS cos θ are given by NCERT. What are they?
E times the projection of the area normal to the field, ΔS cos θ; or the component of E along the normal, E cos θ, times ΔS (NCERT pages 22–23).
Q7. A field E = (2î + 3ĵ + 4k̂) × 103 N C−1 crosses a flat area of 0.02 m2 whose area vector is along +z. Find the flux.
Only the component along the normal counts: ϕ = E·ΔS = (4 × 103)(0.02) = 80 N m2 C−1 (NCERT Eq. (1.11), page 22).
Q8. A hemispherical bowl of radius R sits in a uniform field E along its axis. What is the size of the flux through its curved surface?
πR2E. The bowl and its flat rim form a closed surface, and in a uniform field every line that enters it also leaves, so the net flux is zero (as for the cube in Example 2). The curved surface therefore has the same flux magnitude as the disc of area πR2, with the opposite sign when both normals are outward.