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2 Find the capaci tance ot a parallel plate leapacitor if 2000 uC of change is deposited on its plates when 204 are applied acrass the plates.

2 Find the capaci tance ot a parallel plate leapacitor if 2000 uC of change is deposited on its plates when 204 are applied acrass the plates.

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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**Problem 2**: Find the capacitance of a parallel plate capacitor if 2000 µC of charge is deposited on its plates when 20V are applied across the plates. **Solution Explanation**: To solve for the capacitance \( C \) of a parallel plate capacitor, we use the formula: \[ C = \frac{Q}{V} \] where \( Q \) is the charge stored on the plates, and \( V \) is the voltage applied across the plates. Given: - \( Q = 2000 \, \mu C = 2000 \times 10^{-6} \, C \) - \( V = 20 \, V \) Substitute the given values into the formula: \[ C = \frac{2000 \times 10^{-6} \, \text{C}}{20 \, \text{V}} \] Simplify the expression: \[ C = \frac{2000 \times 10^{-6}}{20} \, \text{F} \] \[ C = \frac{2000}{20} \times 10^{-6} \, \text{F} \] \[ C = 100 \times 10^{-6} \, \text{F} \] \[ C = 100 \, \mu F \] Therefore, the capacitance of the parallel plate capacitor is: \[ \boxed{100 \, \mu F} \] This formula and calculation are fundamental in understanding how capacitance is influenced by the amount of charge stored and the voltage applied across a capacitor's plates. This knowledge is essential for electrical engineering and physics students studying electrical circuits and components.
Transcribed Image Text:**Problem 2**: Find the capacitance of a parallel plate capacitor if 2000 µC of charge is deposited on its plates when 20V are applied across the plates. **Solution Explanation**: To solve for the capacitance \( C \) of a parallel plate capacitor, we use the formula: \[ C = \frac{Q}{V} \] where \( Q \) is the charge stored on the plates, and \( V \) is the voltage applied across the plates. Given: - \( Q = 2000 \, \mu C = 2000 \times 10^{-6} \, C \) - \( V = 20 \, V \) Substitute the given values into the formula: \[ C = \frac{2000 \times 10^{-6} \, \text{C}}{20 \, \text{V}} \] Simplify the expression: \[ C = \frac{2000 \times 10^{-6}}{20} \, \text{F} \] \[ C = \frac{2000}{20} \times 10^{-6} \, \text{F} \] \[ C = 100 \times 10^{-6} \, \text{F} \] \[ C = 100 \, \mu F \] Therefore, the capacitance of the parallel plate capacitor is: \[ \boxed{100 \, \mu F} \] This formula and calculation are fundamental in understanding how capacitance is influenced by the amount of charge stored and the voltage applied across a capacitor's plates. This knowledge is essential for electrical engineering and physics students studying electrical circuits and components.
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