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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

Answer in Brief

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.

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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
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