GATE 2026 IN Q.15: solution and answer

GATE 2026 · INQ.15 · 1 mark · MCQ

An electric field has a potential of $V (x, y, z) = x^{2} + y^{2} + z^{2} V$ .

A charge of 1 coulomb placed at $(\hat{i} + \hat{j} + \hat{k})$ experiences a force of

F = (a \hat{i} + b \hat{j} + c \hat{k}) N

The values of $(a, b, c)$ are ______.

(A)
$a = \frac{1}{3}, b = \frac{1}{3}, c = \frac{1}{3}$
(B)
$a = \frac{1}{3}, b = \frac{1}{3}, c = \frac{1}{3}$
(C)
$a = 0, b = 1, c = \frac{1}{3}$
(D)
$a = 0, b = 1, c = \frac{1}{3}$

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Answer and solution

ANSWER

The force is $F = - \frac{1}{3} (\hat{i} + \hat{j} + \hat{k}) N$ , so strictly $a = b = c = - \frac{1}{3}$ . As printed, no option is exactly correct. Option (A) has the right magnitudes but leaves out the minus sign, which makes it the best match. The official key awards marks to all candidates.

Official answer: marks to all candidatesSource: Official answer key, GATE 2026 IN (IIT Guwahati)

Field from potential. The field is $E = - \nabla V$ . Write $r = x^{2} + y^{2} + z^{2}$ , so $V = r$ .
Gradient of $r$ . $\frac{\partial r}{\partial x} = \frac{x}{r}$ , and likewise for $y$ and $z$ . So $\nabla V = \frac{x i ^ + y j ^ + z k ^}{r} = \overset{a}{^}_{r},$ which is the unit vector pointing radially outward. This holds everywhere except the origin.
Field at the point. The position $\hat{i} + \hat{j} + \hat{k}$ is the point $(1, 1, 1)$ , where $r = 3$ . So $E = - \frac{i ^ + j ^ + k ^}{3} V/m .$ Its magnitude is $∣ E ∣ = 1 V/m$ .
Force on the charge. $F = q E$ with $q = 1 C$ : $F = - \frac{1}{3} \hat{i} - \frac{1}{3} \hat{j} - \frac{1}{3} \hat{k} N,$ so strictly $a = b = c = - \frac{1}{3}$ . Check: the force points toward the origin, where $V$ is lowest. That is right, because a positive charge is pushed toward lower potential.
Matching the options. As printed, no option is exactly correct, because every option gives positive components. Option (A) is the only one with the correct size, $\frac{1}{3}$ , equal on all three axes. It is exactly what you get if the minus sign in $E = - \nabla V$ is dropped. Option (B) gives $\frac{x}{r ^{2}} = \frac{1}{3}$ , dividing by $r^{2}$ instead of by $r$ . Options (C) and (D) give unequal components, which is impossible because both $V$ and the point $(1, 1, 1)$ are symmetric in $x$ , $y$ and $z$ . The organising institute awarded marks to all candidates for this question. The best match is (A), in magnitude only.

KEY IDEA

$E = - \nabla V$ . For $V = r = x^{2} + y^{2} + z^{2}$ , $\nabla r = \overset{a}{^}_{r}$ , so the field has magnitude 1 V/m everywhere and points toward the origin. The force on the charge is $F = q E$ , and at $(1, 1, 1)$ each component is $- \frac{1}{3}$ .

WATCH OUT FOR

There are two. The first is dropping the minus sign in $E = - \nabla V$ , which gives option (A) exactly but puts the force on a positive charge pointing toward higher potential. The second is carrying over the inverse-square habit from a point charge and dividing by $r^{2}$ , which gives (B). Here the potential grows linearly with $r$ , so the field has the same magnitude everywhere; it does not fall off as $1/ r^{2}$ .

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GATE 2026 IN Q.15: solution and answer

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