GATE 2025 IN Q.19: solution and answer

GATE 2025 · INQ.19 · 1 mark · MCQ

An infinite sheet of uniform charge $ρ_{s} = 10 C/m^{2}$ is placed on $z = 0$ plane. The medium surrounding the sheet has a relative permittivity of 10. The electric flux density, in $C/m^{2}$ , at a point $P (0, 0, 5)$ , is

Note: $\overset{a}{^}$ , $\hat{b}$ , and $\overset{c}{^}$ are unit vectors along the $x$ , $y$ , and $z$ directions, respectively.

(A)
$5 \overset{c}{^}$
(B)
$0.25 \overset{c}{^}$
(C)
$10 \overset{c}{^}$
(D)
$0.5 \overset{c}{^}$

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Answer and solution

ANSWER

Option (A): $D = 5 \overset{c}{^} C/m^{2}$ . An infinite sheet gives $D = ρ_{s} /2 = 5 C/m^{2}$ , directed along $+ z$ at $P$ , and this does not depend on the relative permittivity.

Official answer: ASource: Official answer key, GATE 2025 IN (IIT Roorkee)

Gauss's law over a pillbox. Take a cylindrical pillbox of face area $A$ that straddles the sheet. By symmetry $D$ is normal to the sheet and, because the sheet is positively charged, points away from it on both sides. $D$ runs parallel to the curved wall, so that wall carries no flux and only the two flat faces count: $\oint D \cdot d S = D A + D A = Q_{e n c} = ρ_{s} A$
Field of an infinite sheet. So $D = \frac{ρ _{s}}{2}$ . It is the same at every distance from the sheet, so the height $z = 5$ of the point $P$ does not change the magnitude.
The permittivity does not appear. Gauss's law written for $D$ uses only the free charge, so $D$ does not depend on $ϵ_{r}$ . Only $E$ depends on the medium: $E = \frac{ρ _{s}}{2 ϵ _{0} ϵ _{r}} \overset{a}{^}_{n}$ , where $ϵ_{0} = 8.854 \times 1 0^{- 12} F/m$ . Multiplying back, $D = ϵ_{0} ϵ_{r} E = \frac{ρ _{s}}{2} \overset{a}{^}_{n}$ , and the $ϵ_{r} = 10$ cancels.
Magnitude. $D = \frac{10}{2} = 5 C/m^{2}$ .
Direction. $P (0, 0, 5)$ is on the $+ z$ side of the sheet, and the sheet is positively charged, so $D$ at $P$ points away from the sheet, along $+ z$ , which is $\overset{c}{^}$ . Therefore $D = 5 \overset{c}{^} C/m^{2}$ This is option (A).

KEY IDEA

Gauss's law for $D$ uses only the free charge enclosed. For an infinite sheet, $D = \frac{ρ _{s}}{2} \overset{a}{^}_{n}$ whatever the medium and whatever the distance from the sheet. The permittivity changes only $E = D / (ϵ_{0} ϵ_{r})$ .

WATCH OUT FOR

Dividing by $ϵ_{r}$ to get $0.5 \overset{c}{^}$ (option D), or by $2 ϵ_{r}$ again to get $0.25 \overset{c}{^}$ (option B). Both mix up $D$ with $E$ , and only $E$ depends on the medium. A third mistake is to forget that the flux leaves through both faces of the pillbox and write $D = ρ_{s} = 10$ (option C). $D = ρ_{s}$ is the result for a conductor surface, where the field on the inside is zero.

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GATE 2025 IN Q.19: solution and answer

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