Question

A closed triangular box is kept in an electric field of magnitude E = 2 × 10³ N C^-1 as shown in the figure.

Calculate the electric flux through the

1. vertical rectangular surface
2. slanted surface and
3. entire surface.

Answer 1

Electric field of magnitude E = 2 × 10³ NC^-1

(a) Vertical rectangular surface:

Rectangular area A= 5 × 10^-2 × 15 × 10^-2

A= 75 × 10^-24 m²

θ =180°

⇒ cos 180° = -1

Electric flux, Φ_v.s = EA cos θ

= 2 × 10³ × 75 × 10^-4 × cos 180°

= -150 × 10^-1

Φ_v.s = -15 Nm² C^-1

(b) Slanted surface:

cos θ = cos 60° = 0.5

sin θ = sin 30° = `"Opposite"/"hyp"`

hyp = `(5 xx 10^2)/0.5`

hyp = 0.1m

Area of slanted surface A₂ = (0.1 × 15 × 10^-2)

A₂ = 0.015 M²

Electric flux, Φ_v.s = EA = cos θ

= 2 × 10³ × 0.015 × cos 60°

= 2 × 10³ × 0.015 × 10³

= 0.015 × 10³

Φ_v.s = 15 Nm² C^-1

Horizontal surface

θ = 90° ; cos 90° = 0
Electric flux, Φ_H.S = E. A₃ Cos 90° = 0

(c) Entire surface:

Φ_Total = Φ_V.S + Φ_S.S + Φ_H.S = -15 + 15 + 0

Φ_Total = 0