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Anna11 [10]
3 years ago
10

A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structur

e of the capacitor. The charge on the capacitor is held fixed. How is the electric field between the plates of the capacitor affected? A dielectric material, such as Teflon®, is placed between the plates of a parallel-plate capacitor without altering the structure of the capacitor. The charge on the capacitor is held fixed. How is the electric field between the plates of the capacitor affected? The electric field becomes infinite because of the insertion of the Teflon®. The electric field becomes zero after the insertion of the Teflon®. The electric field decreases because of the insertion of the Teflon®. The electric field increases because of the insertion of the Teflon®. The electric field is not altered, because the structure remains unchanged. SubmitRequest Answer Provide Feedback Next
Engineering
1 answer:
Lina20 [59]3 years ago
8 0

Answer: The electric field decreases because of the insertion of the Teflon.

Explanation:

If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.

However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.

In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and  increasing the capacitance proportionally to the dielectric constant of the Teflon.  

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the height = 3 m

(a). the sketch of the thermal circuit is uploaded in the picture below.

(b).  the thermal resistance due to the conduction in the first glass plane is given thus;

R₁ = Lg / Kg A ................(1)

given that Kg rep. the thermal conductivity of the glass plane

A = conduction surface area

Lg = Thickness of glass plane4

taking the thermal conductivity of glass plane as Kg = 0.78 w/mk

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R₁ = [1 (cm) ˣ 1 (m)/100 (cm)] / [(0.78 w/mk)(4m ˣ 3m)]

R₁ = 1.068 ˣ 10 ⁻³ k/w

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therefore R₁ = R₃ = 1.068 ˣ 10 ⁻³ k/w

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R₂ = La/KaA .....................(2)

given Ka = thermal conductivity of air

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substituting values into the equation we have

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q = 557.8 W  

⇒ Applying the heat transfer formula for inside surface glass temperature gives;

q = Ti - T₂ / R₃ + R₄

T₂ = Ti - q (R₃ + R₄)

T₂ = 27 - 557.8 (1.068ˣ10⁻³ + 6.94ˣ10⁻³ ) = 22.55°C

T₂ = 22.55°C

cheers i hope this helps

8 0
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