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Physics for Scientists and Engineers, 6e

Physics for Scientists and Engineers, 6e. Chapter 25 – Electric Potential. In the figure below, two points A and B are located within a region in which there is an electric field. The potential difference Δ V = V B – V A is . positive negative zero.

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Physics for Scientists and Engineers, 6e

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  1. Physics for Scientists and Engineers, 6e Chapter 25 – Electric Potential

  2. In the figure below, two points A and B are located within a region in which there is an electric field. The potential difference ΔV = VB– VAis • positive • negative • zero

  3. When moving straight from A to B, E and ds in Equation 25.3 both point toward the right. Thus, the dot product E · ds is positive and ΔV is negative.

  4. In this figure, a negative charge is placed at A and then moved to B. The change in potential energy of the charge–field system for this process is • positive • negative • zero

  5. From Equation 25.3, ΔU = q0 ΔV, so if a negative test charge is moved through a negative potential difference, the potential energy is positive. Work must be done to move the charge in the direction opposite to the electric force on it.

  6. The labeled points of the figure below are on a series of equipotential surfaces associated with an electric field. Rank (from greatest to least) the work done by the electric field on a positively charged particle that moves along the following transitions. • A -> B, B -> C, C -> D, D -> E • A -> B, D -> E, B -> C, C -> D • B -> C, C -> D, A -> B, D -> E • D -> E, C -> D, B -> C, A -> B

  7. Moving from B to C decreases the electric potential by 2 V, so the electric field performs 2 J of work on each coulomb of positive charge that moves. Moving from C to D decreases the electric potential by 1 V, so 1 J of work is done by the field. It takes no work to move the charge from A to B because the electric potential does not change. Moving from D to E increases the electric potential by 1 V, and thus the field does –1 J of work per unit of positive charge that moves.

  8. For the equipotential surfaces in this figure, what is the approximate direction of the electric field? • Out of the page • Into the page • Toward the right edge of the page • Toward the left edge of the page • Toward the top of the page • Toward the bottom of the page

  9. The electric field points in the direction of decreasing electric potential.

  10. A spherical balloon contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, the electric potential at the surface of the balloon will • increase • decrease • remain the same.

  11. The electric potential is inversely proportional to the radius (see Eq. 25.11).

  12. Recall that the spherical balloon from the last question contains a positively charged object at its center. As the balloon is inflated to a greater volume while the charged object remains at the center, the electric flux through the surface of the balloon will • increase • decrease • remain the same.

  13. Because the same number of field lines passes through a closed surface of any shape or size, the electric flux through the surface remains constant.

  14. In Figure 25.10a, take q1 to be a negative source charge and q2 to be the test charge. If q2 is initially positive and is changed to a charge of the same magnitude but negative, the potential at the position of q2 due to q1 • increases • decreases • remains the same

  15. The potential is established only by the source charge and is independent of the test charge.

  16. Consider the situation from the last question again. When q2 is changed from positive to negative, the potential energy of the two-charge system • increases • decreases • remains the same

  17. The potential energy of the two-charge system is initially negative, due to the products of charges of opposite sign in Equation 25.13. When the sign of q2 is changed, both charges are negative, and the potential energy of the system is positive.

  18. In a certain region of space, the electric potential is zero everywhere along the x axis. From this we can conclude that the x component of the electric field in this region is • zero • in the x direction • in the –x direction.

  19. If the potential is constant (zero in this case), its derivative along this direction is zero.

  20. In a certain region of space, the electric field is zero. From this we can conclude that the electric potential in this region is • zero • constant • positive • negative

  21. If the electric field is zero, there is no change in the electric potential and it must be constant. This constant value could be zero but does not have to be zero.

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