A parallel-plate capacitor with air between its plates has a capacitance of 12\text{ }\mu\… — JAMB UTME

A parallel-plate capacitor with air between its plates has a capacitance of $12\text{ }\mu\text{F}$ . A dielectric slab of relative permittivity $\varepsilon_r = 4$ and thickness $t = \frac{d}{3}$ , where $d$ is the total plate separation, is inserted between the plates parallel to them. What is the effective capacitance of the modified capacitor?

$16\text{ }\mu\text{F}$ Answer
B
$48\text{ }\mu\text{F}$
C
$162\text{ }\mu\text{F}$
D
$6\text{ }\mu\text{F}$

Answer

The effective capacitance of the modified capacitor is

16\text{ }\mu\text{F}

Inserting a dielectric slab of thickness

t = d/3

creates a system equivalent to two series capacitors: an air-filled region of thickness

2d/3

(

C_1 = 1.5 C_0 = 18\text{ }\mu\text{F}

) and a dielectric-filled region of thickness

d/3

(

C_2 = 12 C_0 = 144\text{ }\mu\text{F}

). Combining them via the series reciprocal formula gives

C_{\text{eq}} = \frac{18 \times 144}{18 + 144} = 16\text{ }\mu\text{F}

Step-by-Step Solution

Model the partially filled capacitor as two capacitors connected in series.

Air layer of thickness

d_1 = d - t = \frac{2}{3}d

forms capacitor

C_1

. Dielectric layer of thickness

d_2 = t = \frac{1}{3}d

forms capacitor

C_2

Dividing the plate gap vertically into two distinct media creates two capacitive regions sharing the same electric flux path.

Calculate the individual capacitances

C_1

and

C_2

in terms of initial air capacitance

C_0 = 12\text{ }\mu\text{F}

C_1 = \frac{\varepsilon_0 A}{\frac{2}{3}d} = \frac{3}{2}C_0 = \frac{3}{2}(12) = 18\text{ }\mu\text{F}

and

C_2 = \frac{\varepsilon_r \varepsilon_0 A}{\frac{1}{3}d} = 3 \varepsilon_r C_0 = 3(4)(12) = 144\text{ }\mu\text{F}

Capacitance is inversely proportional to plate distance and directly proportional to relative permittivity.

Calculate the equivalent capacitance

C_{\text{eq}}

for two series capacitors.

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} = \frac{1}{18} + \frac{1}{144} = \frac{8 + 1}{144} = \frac{9}{144} = \frac{1}{16}\text{ }\mu\text{F}^{-1} \implies C_{\text{eq}} = 16\text{ }\mu\text{F}

Capacitors connected in series combine reciprocally.

Key Concept

Partially filled parallel-plate capacitors act as series combinations of distinct capacitive layers.

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