A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d. (a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.) C = (b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.
A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d. (a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.) C = (b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.
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Chapter1: Units, Trigonometry. And Vectors
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A parallel-plate capacitor is constructed by filling the space between two square plates with a block of three dielectric materials, as shown in the figure below. You may assume that ℓ ≫ d.
(a) Find an expression for the capacitance of the device in terms of the plate area A and d, ?1, ?2, and ?3. (Use the following as necessary: ?1, ?2, ?3, ?0, A and d.)
C =
(b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.
C =
(b) Calculate the capacitance using the values A = 1.30 cm2, d = 2.00 mm, ?1 = 4.90, ?2 = 5.60, and ?3 = 2.50.
Transcribed Image Text:d/2
d
K1
K3
e/2-
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