GATE 2025 EE Q.51: solution and answer

GATE 2025 · EEQ.51 · 2 marks · MSQ

Let $a_{R}$ be the unit radial vector in the spherical co-ordinate system. For which of the following value(s) of $n$ , the divergence of the radial vector field $f (R) = a_{R} \frac{1}{R ^{n}}$ is independent of $R$ ?

(A)
$- 2$
(B)
$- 1$
(C)
$1$
(D)
$2$

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Answer and solution

ANSWER

Options (B) $n = - 1$ and (D) $n = 2$ are correct. With $\nabla \cdot f = (2 - n) R^{- (n + 1)}$ , $n = - 1$ gives the constant 3 and $n = 2$ gives 0 everywhere except $R = 0$ .

Official answer: B, DSource: Official answer key, GATE 2025 EE (IIT Roorkee)

The field has only a radial component, $f_{R} = R^{- n}$ , and it does not depend on $θ$ or $ϕ$ . So only the radial term of the spherical divergence survives: $\nabla \cdot f = \frac{1}{R ^{2}} \frac{\partial}{\partial R} (R^{2} f_{R})$
Substitute: $R^{2} f_{R} = R^{2 - n}$ , so $\frac{\partial}{\partial R} (R^{2 - n}) = (2 - n) R^{1 - n}$ .
Divide by $R^{2}$ : $\nabla \cdot f = (2 - n) R^{- (n + 1)}$ This is independent of $R$ only when the power of $R$ is zero or the coefficient is zero.
Case 1, the power of $R$ is zero: $n + 1 = 0 \Rightarrow n = - 1$ . Then $f = R a_{R}$ and $\nabla \cdot f = (2 - (- 1)) R^{0} = 3$ , a constant. So option (B) is correct.
Case 2, the coefficient is zero: $2 - n = 0 \Rightarrow n = 2$ . Then $f = a_{R} / R^{2}$ (the inverse-square field of a point charge) and $\nabla \cdot f = 0$ for every $R \neq = 0$ . Zero is also independent of $R$ , so option (D) is correct.
Check the others: $n = - 2$ gives $\nabla \cdot f = 4 R$ , and $n = 1$ gives $\nabla \cdot f = 1/ R^{2}$ . Both depend on $R$ , so (A) and (C) are wrong. Answer: (B) and (D).

KEY IDEA

For a purely radial field in spherical coordinates, $\nabla \cdot f = \frac{1}{R ^{2}} \frac{d}{d R} (R^{2} f_{R})$ . With $f_{R} = R^{- n}$ this gives $(2 - n) R^{- (n + 1)}$ . That expression can lose its $R$ -dependence in two ways: the power of $R$ becomes zero ( $n = - 1$ ), or the coefficient becomes zero ( $n = 2$ , the inverse-square field, which has zero divergence).

WATCH OUT FOR

Setting only the power of $R$ to zero and marking just (B). Students forget that a divergence of zero is also independent of $R$ , so they miss $n = 2$ , the inverse-square (Coulomb) field. A second trap is using the Cartesian-style derivative $\partial f_{R} / \partial R$ without the $R^{2}$ factor. That gives $- n R^{- n - 1}$ and wrongly picks $n = - 1$ and $n = 0$ .

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GATE 2025 EE Q.51: solution and answer

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