GATE 2018 Electromagnetic Fields Questions and Solutions

Question 01

Question 1

The capacitance of an air-filled parallel--plate capacitor is 60 pF. When a dielectric slab whose thickness is half the distance between the plates is placed on one of the plates, covering it entirely, the capacitance becomes 86 pF. Neglecting the fringing effects, the relative permittivity of the dielectric is ________ (up to 2 decimal places).

Solution

Given:

Original Capacitance (\(C_{\text{air}}\)) = 60 pF.
New Capacitance (\(C_{\text{eq}}\)) = 86 pF.
Dielectric thickness = \(d/2\).

When a dielectric slab of thickness \(d/2\) is inserted, the system can be modeled as two capacitors in series:

\(C_1\): The air-filled portion.
\(C_2\): The dielectric-filled portion.

Step 1: Calculate Individual Capacitances

1. Capacitance of air portion (\(C_1\)): Thickness \(= d/2\), Relative Permittivity \(\varepsilon_r = 1\).

Equation

\[C_1 = \frac{\varepsilon_0 A}{d/2} = \frac{2\varepsilon_0 A}{d} = 2 \times C_{\text{air}} = 2(60 \text{ pF}) = 120 \text{ pF}\]

2. Capacitance of dielectric portion (\(C_2\)): Thickness \(= d/2\), Relative Permittivity \(= \varepsilon_r\).

Equation

\[C_2 = \frac{\varepsilon_0 \varepsilon_r A}{d/2} = \varepsilon_r \left(\frac{2\varepsilon_0 A}{d}\right) = \varepsilon_r (120 \text{ pF}) = 120 \varepsilon_r \text{ pF}\]

Step 2: Calculate Equivalent Capacitance

Since \(C_1\) and \(C_2\) are in series, the equivalent capacitance is:

Equation

\[C_{\text{eq}} = \frac{C_1 C_2}{C_1 + C_2}\]

Substitute the known values (\(C_{\text{eq}} = 86\)):

Equation

\[\begin{aligned} 86 &= \frac{120 \cdot (120 \varepsilon_r)}{120 + 120 \varepsilon_r} \\[10pt] 86 &= \frac{120 \cdot 120 \varepsilon_r}{120(1 + \varepsilon_r)} \\[10pt] 86 &= \frac{120 \varepsilon_r}{1 + \varepsilon_r} \end{aligned}\]

Step 3: Solve for \(\varepsilon_r\)

Equation

\[\begin{aligned} \frac{86}{120} &= \frac{\varepsilon_r}{1 + \varepsilon_r} \\[10pt] 0.7167 (1 + \varepsilon_r) &= \varepsilon_r \quad \text{(approximated ratio)} \\[5pt] \text{Or, using cross-multiplication:} \quad 86(1 + \varepsilon_r) &= 120 \varepsilon_r \\[5pt] 86 + 86 \varepsilon_r &= 120 \varepsilon_r \\[5pt] 86 &= 120 \varepsilon_r - 86 \varepsilon_r \\[5pt] 86 &= 34 \varepsilon_r \\[10pt] \varepsilon_r &= \frac{86}{34} \\[5pt] \varepsilon_r &\approx 2.5294 \end{aligned}\]

Rounding to 2 decimal places:

Equation

\[\boxed{\varepsilon_r = 2.53}\]

✓

Final Answer

Answer: 2.53.