본문 EXERCISES Section 22.1 Electric Potential Difference 16. The potential difference and the work per unit charge, done by an external agent, are equal in magnitude, so (Note: Since only magnitudes are needed in this problem, we omitted the subscripts A and B.) 17. INTERPRET This problem is about the energy gained by an electron as it moves through a potential difference DEVELOP We assume that the electron is initially at rest. When released from the negative plate, it moves toward the positive plate, and the kinetic energy gained is EVALUATE As the electron moves from the negative side to the positive side (i.e., against the direction of the electric field), the kinetic energy it gains is ASSESS Moving a negative charge through a positive potential difference is like going downhill; potential energy decreases. However, the kinetic energy of the electron is increased. 18. The work done by an external agent equals the potential energy change, hence (Since the work required to move the charge from A to B is positive, is positive.) 19. INTERPRET This problem is about conversion of units. DEVELOP By definition, On the other hand, EVALUATE Combining the two expressions gives It follows that ASSESS These are the units for the electric field strength. 20. For in the direction of a uniform electric field, Equation 22.2b gives (See note in solution to Exercise 16. Since the potential always decreases in the direction of the electric field.) 21. INTERPRET This problem is about the work done by the battery to move the charge from the positive terminal to the negative terminal. DEVELOP The work done by the battery is equal to the kinetic energy gained by the charge, and is equal to EVALUATE Substituting the values given, we have ASSESS The charged particle gains kinetic energy as it moves toward the negative plate (in the direction of the electric field). The battery is needed to maintain the potential difference between the plates. 22. The energy gained is (see Example 22.1). The proton and singly-ionized helium atom have charge e, so they gain while the has charge 2e and gains twice this energy. Section 22.2 Calculating Potential Difference 하고 싶은 말 좀 더 업그레이드하여 자료를 보완하여, 과제물을 꼼꼼하게 정성을 들어 작성했습니다. 위 자료 요약정리 잘되어 있으니 잘 참고하시어 학업에 나날이 발전이 있기를 기원합니다 ^^ 구입자 분의 앞날에 항상 무궁한 발전과 행복과 행운이 깃들기를 홧팅 키워드 단원, 핵심, 레포트답안, 전기파트, 핵심물리학, 물리학 |
2017년 5월 26일 금요일
핵심물리학2 전기파트 22단원 레포트답안
핵심물리학2 전기파트 22단원 레포트답안
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