GATE 2020 EE Q.33: solution and answer

GATE 2020 · EEQ.33 · 2 marks · MCQ

The static electric field inside a dielectric medium with relative permittivity, $ε_{r} = 2.25$ , expressed in cylindrical coordinate system is given by the following expression

E = \overset{a}{^}_{r} 2 r + \overset{a}{^}_{φ} (\frac{3}{r}) + \overset{a}{^}_{z} 6

where $\overset{a}{^}_{r}, \overset{a}{^}_{φ}, \overset{a}{^}_{z}$ are unit vectors along $r$ , $φ$ and $z$ directions, respectively. If the above expression represents a valid electrostatic field inside the medium, then the volume charge density associated with this field in terms of free space permittivity, $ε_{0}$ , in SI units is given by:

(A)
$3 ε_{0}$
(B)
$4 ε_{0}$
(C)
$5 ε_{0}$
(D)
$9 ε_{0}$

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Answer and solution

ANSWER

The volume charge density is $ρ_{v} = ε_{0} ε_{r} \nabla \cdot E = 2.25 \times 4 ε_{0} = 9 ε_{0} C/m^{3}$ , option (D).

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

Use Gauss's law in point form. In a linear dielectric $D = ε_{0} ε_{r} E$ , so $ρ_{v} = \nabla \cdot D = ε_{0} ε_{r} \nabla \cdot E$ . The medium is uniform, so $ε_{r}$ comes out of the divergence. Here $ε_{0} \approx 8.854 \times 1 0^{- 12} F/m$ , but the answer is asked for in terms of $ε_{0}$ .
Divergence in cylindrical coordinates: $\nabla \cdot E = \frac{1}{r} \frac{\partial ( r E _{r} )}{\partial r} + \frac{1}{r} \frac{\partial E _{φ}}{\partial φ} + \frac{\partial E _{z}}{\partial z}$ with $E_{r} = 2 r$ , $E_{φ} = 3/ r$ , $E_{z} = 6$ .
Radial term: $r E_{r} = 2 r^{2}$ , so $\frac{1}{r} \frac{\partial ( 2 r ^{2} )}{\partial r} = \frac{4 r}{r} = 4$ .
Azimuthal term: $E_{φ} = 3/ r$ does not depend on $φ$ , so it contributes $0$ . Axial term: $E_{z} = 6$ is constant, so it contributes $0$ . Hence $\nabla \cdot E = 4 V/m^{2}$ .
Check that the field is a valid electrostatic field, i.e. $\nabla \times E = 0$ . In cylindrical coordinates: $r$ -component $\frac{1}{r} \frac{\partial E _{z}}{\partial φ} - \frac{\partial E _{φ}}{\partial z} = 0 - 0 = 0$ ; $φ$ -component $\frac{\partial E _{r}}{\partial z} - \frac{\partial E _{z}}{\partial r} = 0 - 0 = 0$ ; $z$ -component $\frac{1}{r} [\frac{\partial ( r E _{φ} )}{\partial r} - \frac{\partial E _{r}}{\partial φ}] = \frac{1}{r} [\frac{\partial ( 3 )}{\partial r} - 0] = 0$ . So the field is curl-free (for $r \neq = 0$ ), as the question states.
Volume charge density: $ρ_{v} = ε_{0} ε_{r} \nabla \cdot E = 2.25 \times 4 ε_{0} = 9 ε_{0} C/m^{3}$ which is option (D).

KEY IDEA

Free charge comes from the divergence of $D$ , not of $E$ : in a uniform linear dielectric, $ρ_{v} = \nabla \cdot D = ε_{0} ε_{r} \nabla \cdot E$ . In cylindrical coordinates the radial term is $\frac{1}{r} \frac{\partial ( r E _{r} )}{\partial r}$ , and a $φ$ -component that depends only on $r$ adds nothing to the divergence.

WATCH OUT FOR

Using $ρ_{v} = ε_{0} \nabla \cdot E$ and leaving out $ε_{r}$ gives $4 ε_{0}$ (option B), which is the total (free plus bound) charge density, not the free charge. Another slip is to take the radial term as $\partial E_{r} / \partial r = 2$ and forget the $\frac{1}{r} \frac{\partial ( r \cdot )}{\partial r}$ form, which gives $2.25 \times 2 = 4.5 ε_{0}$ , an answer that is not among the options.

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GATE 2020 EE Q.33: solution and answer

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