Determine the capacitances of the capacitors depicted in the figure. The area of the plates is $A$ and the distance between them is $d$ . In each specific part of the figure, the different dielectric slabs have equal thickness, and the dielectric slabs completely fill the space between the plates.

Question

Determine the capacitances of the capacitors depicted in the figure. The area of the plates is

A

and the distance between them is

d

. In each specific part of the figure, the different dielectric slabs have equal thickness, and the dielectric slabs completely fill the space between the plates.
Determine the capacitances of the capacitors depicted in the figure. The area of the plates is A and the distance

Determine the capacitances of the capacitors depicted in the figure. The area of the plates is A and the distance

Quantrex Academy · Accepted Answer

Correct answer: C. (a) (2 K_1 K_2 _0 A)/(d(K_1 + K_2)) (b) (3 K_1 K_2 K_3 _0 A)/(d(K_1 K_2 + K_2 K_3 + K_3 K_1)) (c) ( _0 A(K_1 + K_2))/(2d)

For figure (a), the two dielectric slabs are in series. Each slab has an area

A and thickness

d/2

.<br>The capacitance of the first slab is

C_1 = \dfrac{K_1 \varepsilon_0 A}{d/2} = \dfrac{2 K_1 \varepsilon_0 A}{d}

.<br>The capacitance of the second slab is

C_2 = \dfrac{K_2 \varepsilon_0 A}{d/2} = \dfrac{2 K_2 \varepsilon_0 A}{d}

.<br>The equivalent capacitance

C_a

is given by:<br>

\dfrac{1}{C_a} = \dfrac{1}{C_1} + \dfrac{1}{C_2} = \dfrac{d}{2 K_1 \varepsilon_0 A} + \dfrac{d}{2 K_2 \varepsilon_0 A} = \dfrac{d}{2 \varepsilon_0 A} \left(

\dfrac{K_1 + K_2}{K_1 K_2} \right)<br>

\Rightarrow C_a =

\dfrac{2 K_1 K_2 \varepsilon_0 A}{d(K_1 + K_2)}<br><br>For figure (b), the three dielectric slabs are in series. Each slab has an area

A

and thickness

d/3

.<br>The capacitances are

C_1 =

\dfrac{3 K_1 \varepsilon_0 A}{d},

C_2 =

\dfrac{3 K_2 \varepsilon_0 A}{d}, and

C_3 =

\dfrac{3 K_3 \varepsilon_0 A}{d}.<br>The equivalent capacitance

C_b

is given by:<br>\dfrac{1}{C_b} = \dfrac{1}{C_1} + \dfrac{1}{C_2} + \dfrac{1}{C_3} = \dfrac{d}{3 \varepsilon_0 A} \left( \dfrac{1}{K_1} + \dfrac{1}{K_2} + \dfrac{1}{K_3} \right) = \dfrac{d}{3 \varepsilon_0 A} \left( \dfrac{K_2 K_3 + K_1 K_3 + K_1 K_2}{K_1 K_2 K_3} \right)

\Rightarrow C_b = \dfrac{3 K_1 K_2 K_3 \varepsilon_0 A}{d(K_1 K_2 + K_2 K_3 + K_3 K_1)}

For figure (c), the two dielectric slabs are in parallel. Each slab has an area

A/2

and thickness

d

.
The capacitance of the first slab is

C_1 = \dfrac{K_1 \varepsilon_0 (A/2)}{d} = \dfrac{K_1 \varepsilon_0 A}{2d}

.
The capacitance of the second slab is

C_2 = \dfrac{K_2 \varepsilon_0 (A/2)}{d} = \dfrac{K_2 \varepsilon_0 A}{2d}

.
The equivalent capacitance

C_c

is given by:

C_c = C_1 + C_2 = \dfrac{K_1 \varepsilon_0 A}{2d} + \dfrac{K_2 \varepsilon_0 A}{2d} = \dfrac{\varepsilon_0 A(K_1 + K_2)}{2d}

Determine the capacitances of the capacitors depicted in the...

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