GATE 2022 EC Q.65: solution and answer

GATE 2022 · ECQ.65 · 2 marks · NAT

In an electrostatic field, the electric displacement density vector, $D$ , is given by

D (x, y, z) = (x^{3} i + y^{3} j + x y^{2} k) C/m^{2},

where $i$ , $j$ , $k$ are the unit vectors along $x$ -axis, $y$ -axis, and $z$ -axis, respectively. Consider a cubical region $R$ centered at the origin with each side of length 1 m, and vertices at $(\pm 0.5 m, \pm 0.5 m, \pm 0.5 m)$ . The electric charge enclosed within $R$ is ________ C (rounded off to two decimal places).

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Answer and solution

ANSWER

The charge enclosed within the cube is 0.50 C.

Official answer: 0.48 to 0.52Source: Official answer key, GATE 2022 EC (IIT Kharagpur)

Gauss's law, point form: $\nabla \cdot D = ρ_{v}$ , the volume charge density in $C/m^{3}$ . So the enclosed charge is $Q = \int_{R} (\nabla \cdot D) d V$ , which by the divergence theorem equals the net outward flux $\oint_{S} D \cdot d S$ .
Divergence: $\nabla \cdot D = \frac{\partial ( x ^{3} )}{\partial x} + \frac{\partial ( y ^{3} )}{\partial y} + \frac{\partial ( x y ^{2} )}{\partial z} = 3 x^{2} + 3 y^{2} + 0$ The $z$ -component $x y^{2}$ does not depend on $z$ , so it contributes nothing.
Volume integral over the cube $- 0.5 \leq x, y, z \leq 0.5$ : $\int_{- 0.5}^{0.5} x^{2} d x = \frac{( 0.5 ) ^{3} - ( - 0.5 ) ^{3}}{3} = \frac{0.25}{3} = \frac{1}{12}$ and the same value holds for $\int_{- 0.5}^{0.5} y^{2} d y$ .
Collect the terms: each side has length $1 m$ , so the two remaining integrations each contribute a factor of $1$ : $Q = 3 \cdot \frac{1}{12} \cdot 1 \cdot 1 + 3 \cdot \frac{1}{12} \cdot 1 \cdot 1 = \frac{1}{4} + \frac{1}{4} = 0.5 C$
Check by surface flux: on the faces $x = \pm 0.5$ , $D_{x} = x^{3} = \pm 0.125$ and the outward normal is $\pm \overset{a}{^}_{x}$ , so $D \cdot \overset{n}{^} = + 0.125$ on both. Each face has area $1 m^{2}$ , so the pair gives $0.25 C$ . The faces $y = \pm 0.5$ give another $0.25 C$ in the same way. On the faces $z = \pm 0.5$ , $D_{z} = x y^{2}$ is the same on both faces but the outward normals point in opposite directions, so those two fluxes cancel. Total $= 0.50 C$ , which agrees with the volume integral.

KEY IDEA

Enclosed charge is the volume integral of $\nabla \cdot D$ (Gauss's law in point form plus the divergence theorem). Here $\nabla \cdot D = 3 x^{2} + 3 y^{2}$ . Over a cube centred at the origin, $\int_{- 0.5}^{0.5} x^{2} d x = 1/12$ , so each term gives $1/4$ and $Q = 0.5 C$ .

WATCH OUT FOR

Integrating from 0 to 1 instead of from $- 0.5$ to $0.5$ . That gives $\int_{0}^{1} x^{2} d x = 1/3$ and $Q = 2 C$ , because the cube is centred at the origin, not placed at a corner. Other slips: differentiating $x y^{2}$ with respect to $x$ or $y$ instead of $z$ , which adds a spurious term, or using $\int_{- 0.5}^{0.5} x^{2} d x = 2 (0.5)^{3} = 0.25$ and forgetting to divide by 3.

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