GATE 2024 EC Q.16: solution and answer

GATE 2024 · ECQ.16 · 1 mark · MCQ

Let $\hat{i}$ and $\hat{j}$ be the unit vectors along $x$ and $y$ axes, respectively and let $A$ be a positive constant. Which one of the following statements is true for the vector fields $F_{1} = A (\hat{i} y + \hat{j} x)$ and $F_{2} = A (\hat{i} y - \hat{j} x)$ ?

(A)
Both $F_{1}$ and $F_{2}$ are electrostatic fields.
(B)
Only $F_{1}$ is an electrostatic field.
(C)
Only $F_{2}$ is an electrostatic field.
(D)
Neither $F_{1}$ nor $F_{2}$ is an electrostatic field.

See the question in the paper

Tap to open full size

Answer and solution

ANSWER

(B) Only $F_{1}$ is an electrostatic field: $\nabla \times F_{1} = 0$ (with $V = - A x y$ ), while $\nabla \times F_{2} = - 2 A \hat{k} \neq = 0$ .

Official answer: BSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

A static electric field is conservative. It must satisfy $\nabla \times E = 0$ , which is the same as saying $E = - \nabla V$ for some scalar potential $V$ . So we check the curl of each field.
Both fields lie in the $x y$ -plane and depend only on $x$ and $y$ , so only the $z$ -component of the curl can be non-zero: $(\nabla \times F)_{z} = \frac{\partial F _{y}}{\partial x} - \frac{\partial F _{x}}{\partial y}$
For $F_{1}$ : $F_{x} = A y$ and $F_{y} = A x$ , so $(\nabla \times F_{1})_{z} = \frac{\partial ( A x )}{\partial x} - \frac{\partial ( A y )}{\partial y} = A - A = 0$ $F_{1}$ is irrotational, so it can be an electrostatic field.
Confirm with a potential. Take $V = - A x y$ . Then $- \nabla V = A y \hat{i} + A x \hat{j} = F_{1}$ , so $F_{1}$ really is the negative gradient of a scalar potential. (Also $\nabla^{2} V = 0$ , so this is the field of a charge-free region.)
For $F_{2}$ : $F_{x} = A y$ and $F_{y} = - A x$ , so $(\nabla \times F_{2})_{z} = \frac{\partial ( - A x )}{\partial x} - \frac{\partial ( A y )}{\partial y} = - A - A = - 2 A \neq = 0$ because $A > 0$ .
$F_{2}$ has non-zero curl, so no scalar potential exists for it. Physically, $F_{2} \cdot (x \hat{i} + y \hat{j}) = A (x y - x y) = 0$ : the field is everywhere perpendicular to the position vector, so its lines are circles about the origin (a vortex field). A static electric field cannot circulate like this, so $F_{2}$ is not electrostatic.
Only $F_{1}$ is an electrostatic field. The answer is (B).

KEY IDEA

A vector field can be an electrostatic field exactly when it is conservative, meaning $\nabla \times E = 0$ (equivalently $E = - \nabla V$ ). The divergence imposes no condition. It only fixes the charge density through $\nabla \cdot E = ρ_{v} / ϵ_{0}$ .

WATCH OUT FOR

Testing the divergence instead of the curl. Both fields have zero divergence, so a divergence test would wrongly pick (A). The other common slip is a sign error in $\partial F_{y} / \partial x$ for $F_{2}$ : its $F_{y} = - A x$ , so the curl is $- 2 A$ , not zero.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 23 Electrostatics questions, with notes

GATE 2024 EC Q.16: solution and answer

Answer and solution

More Electrostatics questions