# Find the equivalent capacitance between P and Q in the following diagram. The area of each plate is A and the separation between plates is d. - Physics

Find the equivalent capacitance between P and Q in the following diagram. The area of each plate is A and the separation between plates is d.

#### Solution

(i) The capacitor in the first diagram is a series combination of three capacitors of plate separations d/3 and plate areas A, with C1 filled with air (k1= 1), C2 filled with a dielectric of k2 = 3, and C3 filled with a dielectric of k3 = 6.

∴ "C"_1 = ("k"_1ε_0"A")/("d"//3) = (3ε_0"A")/"d" "k"_1, "C"_2 = ("k"_2ε_0"A")/("d"//3) = (3ε_0"A")/"d" "k"_2,

"C"_3 = ("k"_3ε_0"A")/("d"//3) = (3ε_0"A")/"d" "k"_3

∴ 1/"C'" = 1/"C"_1 + 1/"C"_2 + 1/"C"_3 = "d"/(3ε_0"A")(1/"k"_1 + 1/"k"_2 + 1/"k"_3)

∴ C' = (3ε_0"A")/"d"(("k"_1"k"_2"k"_3)/("k"_1"k"_2 + "k"_2"k"_3 + "k"_3"k"_1)) = (3ε_0"A")/"d" (18/27)

= (2ε_0"A")/"d"

(ii) The above second figure, a series combination of two capacitors C2 (k2 = 3) and C3 (k3 = 6), of plate areas A/2 and plate separations d/2, is in parallel with a capacitor C1 (k1 =4) of plate area A/2 and plate separation d.

∴ "C"_1 = ("k"_1ε_0("A"//2))/"d" = (ε_0"A")/"2d" "k"_1

"C"_2 = ("k"_2ε_0("A"//2))/("d"//2) = (ε_0"A")/"d" "k"_2

"C"_3 = ("k"_3ε_0("A"//2))/("d"//2) = (ε_0"A")/"d" "k"_3

∴ For the series combination of C2 and C3,

1/"C"_4 = 1/"C"_2 + 1/"C"_3

∴ "C"_4 = ("C"_2"C"_3)/("C"_2 + "C"_3) = (ε_0"A")/"d"(("k"_2"k"_3)/("k"_2 + "k"_3)) = (ε_0"A")/"d" ((3xx6)/(3+6)) = (2ε_0"A")/"d"

Finally, for the parallel combination of C1 and C4

C'' = "C"_1 + "C"_4 = (ε_0"A")/"2d" (4) + (2ε_0"A")/"d" = (4ε_0"A")/"d"

Thus, the equivalent capacitances are C' = (2ε_0"A")/"d" and C'' = (4ε_0"A")/"d".

Concept: Capacitors and Capacitance, Combination of Capacitors in Series and Parallel
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#### APPEARS IN

Balbharati Physics 12th Standard HSC Maharashtra State Board
Chapter 8 Electrostatics
Exercises | Q 13 | Page 213