The total charge enclosed in an incremental volume of 2 × 10⁻⁹ m³ located at the origin is ___________ nC, if electric flux density of its field is found as
D = e⁻ˣ sin y î - e⁻ˣ cos y ĵ + 2z k̂ C/m²

Solution

Physics sir · 4 steps

Gauss's law in point form: the divergence of D gives charge per unit volume, then multiply by the tiny volume.

1. Point form. Gauss's law for a very small volume says that the divergence of D⃗ equals the volume charge density ρ. Then the enclosed charge is Q = ρ Δ V.

   ∇ · D⃗ = (∂ Dₓ)/(∂ x) + (∂ D_y)/(∂ y) + (∂ D_z)/(∂ z) = ρ

2. Take each derivative. ∂/(∂ x)(e⁻ˣ sin y) = -e⁻ˣ sin y and ∂/(∂ y)(-e⁻ˣ cos y) = +e⁻ˣ sin y. These two cancel, and ∂/(∂ z)(2z) = 2. So ρ = 2 C m⁻³ at every point, including the origin.
3. Multiply by volume. Q = 2 × 2 × 10⁻⁹ = 4 × 10⁻⁹ C.
4. So. The enclosed charge is 4 × 10⁻⁹ C, which is 4 nC.

The common slip: 2. Stopping at ∇ · D⃗ gives 2, but that is the charge density in C m⁻³. You still have to multiply by the volume 2 × 10⁻⁹ m³.

The idea to keep. For a tiny volume, do not integrate flux over a surface. Find ∇ · D⃗ at that point to get ρ, then multiply by the volume to get the charge.

Answer from the NTA final answer key. Solution by PrepWiser.