**Electric Field Strength Calculation****Problem Statement:**A double charged ion is accelerated to an energy of 30.9 keV by the electric field between two parallel conducting plates separated by 2.4 cm. What is the electric field strength between the plates?**Solution:**To find the electric field (\( E \)) between the plates, we need to use the relationship between the energy (\( U \)), charge (\( q \)), and the separation (\( d \)):\[ U = qEd \]Given:- \( U \) = 30.9 keV = \( 30.9 \times 10^3 \) eV- The ion is double charged, so \( q = 2e \), where \( e \) is the charge of an electron (approximately \( 1.602 \times 10^{-19} \) C)- \( d \) = 2.4 cm = 0.024 mSolving for \( E \):\[ E = \frac{U}{qd} = \frac{30.9 \times 10^3 \text{ eV}}{2e \times 0.024 \text{ m}} \]Convert 30.9 keV to Joules (since 1 eV = \( 1.602 \times 10^{-19} \) J):\[ 30.9 \times 10^3 \text{ eV} = 30.9 \times 10^3 \times 1.602 \times 10^{-19} \text{ J} = 4.949 \times 10^{-15} \text{ J} \]Then, substitute \( U = 4.949 \times 10^{-15} \text{ J} \):\[ E = \frac{4.949 \times 10^{-15} \text{ J}}{2 \times 1.602 \times 10^{-19} \text{ C} \times 0.024 \text{ m}} = \frac{4.949 \times 10^{-15}}{7.697 \times 10^{-21}} \]\[ E = 6.429 \times 10^5 \text{ V/m} \]Therefore, the electric field strength between the plates is:\[ E = 6.429 \times 10^5

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A double charged ion is accelerated to an energy of 30.9 keV by the electric field between two parallel conducting plates separated by 2.4 cm. What is the electric field strength between the plates?

Physics for Scientists and Engineers with Modern Physics

10th Edition

ISBN:9781337553292

Author:Raymond A. Serway, John W. Jewett

Publisher:Raymond A. Serway, John W. Jewett

Chapter22: Electric Fields

Section: Chapter Questions

Problem 42AP

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