GATE 2020 EE Q.40: solution and answer

GATE 2020 · EEQ.40 · 2 marks · MCQ

Let $a_{r}$ , $a_{ϕ}$ and $a_{z}$ be unit vectors along $r$ , $ϕ$ and $z$ directions, respectively in the cylindrical coordinate system. For the electric flux density given by $D = (a_{r} 15 + a_{ϕ} 2 r - a_{z} 3 r z) Coulomb/m^{2}$ , the total electric flux, in Coulomb, emanating from the volume enclosed by a solid cylinder of radius 3 m and height 5 m oriented along the z-axis with its base at the origin is:

(A)
$54 π$
(B)
$90 π$
(C)
$108 π$
(D)
$180 π$

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Answer and solution

ANSWER

The total electric flux leaving the cylinder is $180 π C$ , which is option (D).

Official answer: DSource: Official answer key, GATE 2020 EE (IIT Delhi)

The total flux leaving a closed surface is $Ψ = \oint_{S} D \cdot d S = \int_{V} (\nabla \cdot D) d V$ (divergence theorem). The volume is $0 \leq r \leq 3$ , $0 \leq ϕ \leq 2 π$ , $0 \leq z \leq 5$ .
Here $D_{r} = 15$ , $D_{ϕ} = 2 r$ , $D_{z} = - 3 r z$ . The divergence in cylindrical coordinates is $\nabla \cdot D = \frac{1}{r} \frac{\partial ( r D _{r} )}{\partial r} + \frac{1}{r} \frac{\partial D _{ϕ}}{\partial ϕ} + \frac{\partial D _{z}}{\partial z} = \frac{1}{r} \frac{\partial ( 15 r )}{\partial r} + 0 + \frac{\partial ( - 3 r z )}{\partial z} = \frac{15}{r} - 3 r$ The $ϕ$ term is zero because $D_{ϕ} = 2 r$ does not depend on $ϕ$ .
The volume element is $d V = r d r d ϕ d z$ , so the integrand becomes $(\frac{15}{r} - 3 r) r = 15 - 3 r^{2}$ .
$Ψ = \int_{0}^{5} \int_{0}^{2 π} \int_{0}^{3} (15 - 3 r^{2}) d r d ϕ d z = [15 r - r^{3}]_{0}^{3} \times 2 π \times 5 = (45 - 27) (10 π) = 180 π C$
Check by integrating over the three faces. Curved side ( $r = 3$ , outward normal $\overset{a}{^}_{r}$ , $D_{r} = 15$ ): $15 \times (2 π \cdot 3 \cdot 5) = 450 π$ . Top ( $z = 5$ , outward normal $\overset{a}{^}_{z}$ , $D_{z} = - 15 r$ ): $\int_{0}^{2 π} \int_{0}^{3} (- 15 r) r d r d ϕ = - 15 \cdot \frac{27}{3} \cdot 2 π = - 270 π$ . Bottom ( $z = 0$ ): $D_{z} = 0$ , so the flux is $0$ . The $\overset{a}{^}_{ϕ}$ component is tangential to every face and contributes nothing.
Total: $450 π - 270 π + 0 = 180 π C$ , which agrees with the volume integral. The answer is option (D).

KEY IDEA

The flux out of a closed surface is the volume integral of $\nabla \cdot D$ (divergence theorem). In cylindrical coordinates the radial term is $\frac{1}{r} \frac{\partial ( r D _{r} )}{\partial r}$ , and the extra $r$ in $d V = r d r d ϕ d z$ cancels the $\frac{1}{r}$ . The $ϕ$ component carries no flux through any face of a cylinder centred on the z-axis.

WATCH OUT FOR

Taking $\frac{\partial D _{r}}{\partial r} = 0$ (as if in Cartesian coordinates), which drops the radial term and gives $- 270 π$ , which is not an option. Leaving out the Jacobian $r$ in $d V$ is another route to a wrong answer. Other slips are giving the top face the wrong sign (its outward normal is $+ \overset{a}{^}_{z}$ and $D_{z} = - 15 r$ there, so its flux is negative) and counting flux from $D_{ϕ}$ , which is tangential to every face.

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GATE 2020 EE Q.40: solution and answer

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