GATE 2026 IN Q.37: solution and answer

GATE 2026 · INQ.37 · 2 marks · MCQ

An electric field in free space is

E = (2 x + 5 y + 6 z) \hat{i} + (5 x + 4 y + 10 z) \hat{j} + (6 x + 10 y + 2 z) \hat{k} V/m .

The charge density is ____ $C/m^{3}$ . ( $ϵ_{0}$ is the permittivity of free space)

(A)
$8 ϵ_{0}$
(B)
$10 ϵ_{0}$
(C)
$6 ϵ_{0}$
(D)
$4 ϵ_{0}$

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Answer and solution

ANSWER

The answer is (A), $8 ϵ_{0} C/m^{3}$ . Here $\nabla \cdot E = 2 + 4 + 2 = 8$ , and $ρ_{v} = ϵ_{0} \nabla \cdot E$ .

Official answer: ASource: Official answer key, GATE 2026 IN (IIT Guwahati)

Use the point form of Gauss's law in free space: $\nabla \cdot D = ρ_{v}$ with $D = ϵ_{0} E$ , so $ρ_{v} = ϵ_{0} \nabla \cdot E$ , where $ϵ_{0} \approx 8.854 \times 1 0^{- 12} F/m$ .
The divergence uses only the diagonal derivatives, each component differentiated with respect to its own coordinate: $\nabla \cdot E = \frac{\partial E _{x}}{\partial x} + \frac{\partial E _{y}}{\partial y} + \frac{\partial E _{z}}{\partial z}$ .
Compute each term: $\frac{\partial}{\partial x} (2 x + 5 y + 6 z) = 2$ , $\frac{\partial}{\partial y} (5 x + 4 y + 10 z) = 4$ , $\frac{\partial}{\partial z} (6 x + 10 y + 2 z) = 2$ .
So $\nabla \cdot E = 2 + 4 + 2 = 8 V/m^{2}$ . It is a constant, so the charge density is the same at every point.
Check that the field is a valid electrostatic field ( $\nabla \times E = 0$ ): $\frac{\partial E _{z}}{\partial y} - \frac{\partial E _{y}}{\partial z} = 10 - 10 = 0$ , $\frac{\partial E _{x}}{\partial z} - \frac{\partial E _{z}}{\partial x} = 6 - 6 = 0$ , $\frac{\partial E _{y}}{\partial x} - \frac{\partial E _{x}}{\partial y} = 5 - 5 = 0$ . The curl is zero, so the field is conservative, as a static field has to be.
Therefore $ρ_{v} = ϵ_{0} (8) = 8 ϵ_{0} C/m^{3},$ which is option (A).

KEY IDEA

In free space, the point form of Gauss's law gives $ρ_{v} = ϵ_{0} \nabla \cdot E$ . Only the diagonal terms count: $\partial E_{x} / \partial x$ , $\partial E_{y} / \partial y$ and $\partial E_{z} / \partial z$ . The off-diagonal coefficients (5, 6, 10) do not affect the charge density. Here they are symmetric, which is exactly the condition for $\nabla \times E = 0$ .

WATCH OUT FOR

Adding every coefficient, or picking up the off-diagonal ones (for example $\partial E_{y} / \partial z = 10$ ), gives distractors such as $10 ϵ_{0}$ . The divergence takes only the $x$ -derivative of $E_{x}$ , the $y$ -derivative of $E_{y}$ and the $z$ -derivative of $E_{z}$ . Another slip is dividing by $ϵ_{0}$ instead of multiplying: $ρ_{v} = ϵ_{0} \nabla \cdot E$ , not $\nabla \cdot E / ϵ_{0}$ .

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GATE 2026 IN Q.37: solution and answer

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