Question

In an electrostatic field, the displacement vector is \[ \vec{D}(x,y,z) = (x^3\,\vec{i} + y^3\,\vec{j} + xy^2\,\vec{k})\ \text{C/m}^2. \] A cube of side 1 m is centered at the origin with vertices at \((\pm 0.5, \pm 0.5, \pm 0.5)\). Find the electric charge enclosed within the cube (rounded to two decimals).

Hint:

For symmetric regions, exploit symmetry to reduce triple integrals—here $x^2$ and $y^2$ contribute equally.

Answer 1

Correct Answer: 0.48

Using Gauss’s law: \[ Q = \iiint_R (\nabla \cdot \vec{D}) \, dV \] Compute divergence: \[ \nabla \cdot \vec{D} = \frac{\partial}{\partial x}(x^3) + \frac{\partial}{\partial y}(y^3) + \frac{\partial}{\partial z}(xy^2) \] \[ \nabla \cdot \vec{D} = 3x^2 + 3y^2 + 0. \] Thus: \[ Q = \iiint_R (3x^2 + 3y^2)\, dV \] Because the cube is symmetric: \[ Q = 3\iiint x^2\, dV + 3\iiint y^2\, dV \] Both integrals are equal, so: \[ Q = 6 \iiint_R x^2\, dV \] Compute: \[ \iiint_R x^2\, dV = \left( \int_{-0.5}^{0.5} x^2\, dx \right) \left( \int_{-0.5}^{0.5} dy \right) \left( \int_{-0.5}^{0.5} dz \right) \] \[ \int_{-0.5}^{0.5} x^2\, dx = 2\int_{0}^{0.5} x^2\, dx = 2\left[\frac{x^3}{3}\right]_{0}^{0.5} = 2 \cdot \frac{0.125}{3} = \frac{0.25}{3} = 0.08333 \] The remaining integrals give: \[ \int_{-0.5}^{0.5} dy = 1,\quad \int_{-0.5}^{0.5} dz = 1 \] Thus: \[ \iiint_R x^2\, dV = 0.08333 \] So: \[ Q = 6(0.08333) = 0.49998\ \text{C} \] Rounded to two decimals: \[ \boxed{0.50\ \text{C}} \]