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What happens to the capacity of a capacitor when its plate distance is doubled, while its dielectric constant is increased by 10%? C2 = .55 C1 O C1 = .55 C2 O The capacity does not change O C2 = 1.8 C1 C2 = 0.05 C1

What happens to the capacity of a capacitor when its plate distance is doubled, while its dielectric constant is increased by 10%? C2 = .55 C1 O C1 = .55 C2 O The capacity does not change O C2 = 1.8 C1 C2 = 0.05 C1

Introductory Circuit Analysis (13th Edition)
Introductory Circuit Analysis (13th Edition)
13th Edition
ISBN:9780133923605
Author:Robert L. Boylestad
Publisher:Robert L. Boylestad
Chapter1: Introduction
Section: Chapter Questions
Problem 1P: Visit your local library (at school or home) and describe the extent to which it provides literature...
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**Question:** What happens to the capacity of a capacitor when its plate distance is doubled, while its dielectric constant is increased by 10%? **Options:** - C2 = 0.55 C1 (Selected) - C1 = 0.55 C2 - The capacity does not change - C2 = 1.8 C1 - C2 = 0.05 C1 **Explanation:** The question asks about the effect on the capacitance of a capacitor when the distance between its plates is doubled and its dielectric constant is increased by 10%. The selected answer indicates that the new capacitance, C2, is 0.55 times the original capacitance, C1. To understand the problem, remember that the capacitance (C) of a parallel plate capacitor is given by the formula: \[ C = \frac{{\varepsilon \cdot A}}{{d}} \] Where: - \( \varepsilon \) is the permittivity of the dielectric material between the plates, which is affected by the dielectric constant (k). - \( A \) is the area of one of the plates. - \( d \) is the distance between the plates. When the plate distance is doubled (\( d = 2d \)), the capacitance formula becomes: \[ C2 = \frac{{\varepsilon' \cdot A}}{{2d}} \] Where \( \varepsilon' \) is 1.1 times the original permittivity (\( 1.1 \varepsilon \)) because of the 10% increase in the dielectric constant: \[ C2 = \frac{{1.1 \varepsilon \cdot A}}{{2d}} = \frac{{1.1}}{2} \cdot \frac{{\varepsilon \cdot A}}{{d}} = \frac{1.1}{2} \cdot C1 = 0.55 C1 \] This confirms that C2 is 0.55 times C1, which matches the selected answer.
Transcribed Image Text:**Question:** What happens to the capacity of a capacitor when its plate distance is doubled, while its dielectric constant is increased by 10%? **Options:** - C2 = 0.55 C1 (Selected) - C1 = 0.55 C2 - The capacity does not change - C2 = 1.8 C1 - C2 = 0.05 C1 **Explanation:** The question asks about the effect on the capacitance of a capacitor when the distance between its plates is doubled and its dielectric constant is increased by 10%. The selected answer indicates that the new capacitance, C2, is 0.55 times the original capacitance, C1. To understand the problem, remember that the capacitance (C) of a parallel plate capacitor is given by the formula: \[ C = \frac{{\varepsilon \cdot A}}{{d}} \] Where: - \( \varepsilon \) is the permittivity of the dielectric material between the plates, which is affected by the dielectric constant (k). - \( A \) is the area of one of the plates. - \( d \) is the distance between the plates. When the plate distance is doubled (\( d = 2d \)), the capacitance formula becomes: \[ C2 = \frac{{\varepsilon' \cdot A}}{{2d}} \] Where \( \varepsilon' \) is 1.1 times the original permittivity (\( 1.1 \varepsilon \)) because of the 10% increase in the dielectric constant: \[ C2 = \frac{{1.1 \varepsilon \cdot A}}{{2d}} = \frac{{1.1}}{2} \cdot \frac{{\varepsilon \cdot A}}{{d}} = \frac{1.1}{2} \cdot C1 = 0.55 C1 \] This confirms that C2 is 0.55 times C1, which matches the selected answer.
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