42. Let C

1

= ε

0

(A/2)κ

1

/2d = ε

0

Aκ

1

/4d, C

2

= ε

0

(A/2)κ

2

/d = ε

0

Aκ

2

/2d, and C

3

= ε

0

Aκ

3

/2d. Note that

C

2

and C

3

are effectively connected in series, while C

1

is effectively connected in parallel with the C

2

-C

3

combination. Thus,

C

=

C

1

+

C

2

C

3

C

2

+ C

3

=

ε

0

Aκ

1

4d

+

(ε

0

A/d)(κ

2

/2)(κ

3

/2)

κ

2

/2 + κ

3

/2

=

ε

0

A

4d

κ

1

+

2κ

2

κ

3

κ

2

+ κ

3



.