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54. (II) How much energy is stored by the electric field between two square plates, 8.0 cm on a side, separated by a 1.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 370 μC.

54. (II) How much energy is stored by the electric field between two square plates, 8.0 cm on a side, separated by a 1.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 370 μC.

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
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**Problem 54: Calculation of Energy Stored in an Electric Field Between Two Plates** **Problem Statement:** How much energy is stored by the electric field between two square plates, 8.0 cm on a side, separated by a 1.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 370 μC. **Given Data:** - Side length of the square plates, \( a = 8.0 \) cm = 0.08 m - Separation between the plates, \( d = 1.5 \) mm = 0.0015 m - Magnitude of charge on each plate, \( Q = 370 \) μC = \( 370 \times 10^{-6} \) C To find the amount of energy stored in the electric field between the plates, we use the formula for the energy stored in a capacitor: \[ U = \frac{1}{2} \frac{Q^2}{C} \] Where \( C \) is the capacitance of the parallel plate capacitor, given by: \[ C = \frac{\epsilon_0 A}{d} \] Here, \( \epsilon_0 \) (the permittivity of free space) is \( 8.85 \times 10^{-12} \) F/m and \( A \) (the area of the plates) is: \[ A = a^2 = (0.08)^2 = 0.0064 \, \text{m}^2 \] Substituting \( A \) and \( d \) into the capacitance formula: \[ C = \frac{8.85 \times 10^{-12} \times 0.0064}{0.0015} \] \[ C = 3.776 \times 10^{-13} \, \text{F} \] Finally, substituting \( C \) and \( Q \) into the energy formula: \[ U = \frac{1}{2} \frac{(370 \times 10^{-6})^2}{3.776 \times 10^{-13}} \] \[ U = \frac{1}{2} \frac{136900 \times 10^{-12}}{3.776 \times 10^{-13}} \] \[ U = \frac{1}{2} \times 362632.98 \
Transcribed Image Text:**Problem 54: Calculation of Energy Stored in an Electric Field Between Two Plates** **Problem Statement:** How much energy is stored by the electric field between two square plates, 8.0 cm on a side, separated by a 1.5-mm air gap? The charges on the plates are equal and opposite and of magnitude 370 μC. **Given Data:** - Side length of the square plates, \( a = 8.0 \) cm = 0.08 m - Separation between the plates, \( d = 1.5 \) mm = 0.0015 m - Magnitude of charge on each plate, \( Q = 370 \) μC = \( 370 \times 10^{-6} \) C To find the amount of energy stored in the electric field between the plates, we use the formula for the energy stored in a capacitor: \[ U = \frac{1}{2} \frac{Q^2}{C} \] Where \( C \) is the capacitance of the parallel plate capacitor, given by: \[ C = \frac{\epsilon_0 A}{d} \] Here, \( \epsilon_0 \) (the permittivity of free space) is \( 8.85 \times 10^{-12} \) F/m and \( A \) (the area of the plates) is: \[ A = a^2 = (0.08)^2 = 0.0064 \, \text{m}^2 \] Substituting \( A \) and \( d \) into the capacitance formula: \[ C = \frac{8.85 \times 10^{-12} \times 0.0064}{0.0015} \] \[ C = 3.776 \times 10^{-13} \, \text{F} \] Finally, substituting \( C \) and \( Q \) into the energy formula: \[ U = \frac{1}{2} \frac{(370 \times 10^{-6})^2}{3.776 \times 10^{-13}} \] \[ U = \frac{1}{2} \frac{136900 \times 10^{-12}}{3.776 \times 10^{-13}} \] \[ U = \frac{1}{2} \times 362632.98 \
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