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Two parallel plate capacitors have identical plate areas and identical plate separations. The maximum energy each can store is determined by the maximum potential difference that can be applied before dielectric breakdown occurs. One capacitor has air between its plates, and the other has Mylar. Find the ratio of the maximum energy the Mylar capacitor can store to the maximum energy the air capacitor can store.

Short Answer

Expert verified
Answer: U_ratio = (3.1 * V_1^2) / (V_0^2), where V_0 and V_1 are the breakdown voltages for air and Mylar capacitors, respectively. To obtain the numerical value of this ratio, the actual breakdown voltages V_0 and V_1 need to be found through research or given data.

Step by step solution

01

Determine the dielectric constants and breakdown voltages

First, we need to find the dielectric constants (K) and breakdown voltages (V) for air and Mylar. The dielectric constants can be found in a reference table or online. The breakdown voltages can also be found using a reference table or can be calculated using the formula: V_breakdown = K * E_breakdown where E_breakdown is the electronegativity of the dielectric material. For simplicity, we will assume that the breakdown voltages are given, but you could find the values through research if desired. According to available resources, we find: - For air: K_air = 1, V_air = V_0 (let's assume its breakdown voltage is V_0) - For Mylar: K_mylar = 3.1, V_mylar = V_1 (let's assume its breakdown voltage is V_1)
02

Compute the capacitance for each capacitor

To calculate the capacitance, we need to use the formula for capacitance for parallel plate capacitors: C = (ε_0 * K * A) / d where ε_0 is the vacuum permittivity, K is the dielectric constant, A is the plate area, and d is the plate separation. Since both capacitors have identical plate areas and plate separations, we only need to adjust the dielectric constant for each capacitor: C_air = (ε_0 * K_air * A) / d = (ε_0 * 1 * A) /d = ε_0 * A/d C_mylar = (ε_0 * K_mylar * A) / d = (ε_0 * 3.1 * A) /d = 3.1 * ε_0 * A/d
03

Compute the maximum energy for each capacitor

To find the maximum energy each capacitor can store, we need to use the formula for the energy stored in a capacitor: U = 0.5 * C * V^2 Using the calculated capacitances and breakdown voltages for each capacitor: U_air = 0.5 * C_air * V_0^2 = 0.5 * (ε_0 * A/d) * V_0^2 U_mylar = 0.5 * C_mylar * V_1^2 = 0.5 * (3.1 * ε_0 * A/d) * V_1^2
04

Compute the ratio of maximum energy storage between the capacitors

We are asked to find the ratio of maximum energy stored in the Mylar capacitor to the maximum energy stored in the air capacitor. To do this, we will divide the energy stored in the Mylar capacitor by the energy stored in the air capacitor: U_ratio = U_mylar / U_air U_ratio = [(0.5 * (3.1 * ε_0 * A/d) * V_1^2)] / [(0.5 * (ε_0 * A/d) * V_0^2)] Notice that the factors 0.5, ε_0, A, and d all cancel out: U_ratio = (3.1 * V_1^2) / (V_0^2) This is our final answer for the ratio of maximum energy stored in the Mylar capacitor to the maximum energy stored in the air capacitor. To obtain the numerical value of this ratio, we would need the actual breakdown voltages V_0 and V_1 for air and Mylar capacitors, which can be found through research or from given data.

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Most popular questions from this chapter

The battery of an electric car stores 60.51 MJ of energy. If 6990 supercapacitors, each with capacitance \(C=3.423 \mathrm{kF}\), are required to supply this amount of energy, what is the potential difference across each supercapacitor?

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The battery of an electric car stores 67.39 MJ of energy. If 6845 supercapacitors, each with capacitance \(C\) and charged to a potential difference of \(2.377 \mathrm{~V}\), can supply this amount of energy, what is the value of \(C\) for each supercapacitor?

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A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

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