Chapter 20: Electric Potential and Electric Potential EnergyAnswers to Even-Numbered Conceptual Questions2. The two like charges, if released, will move away from one another to infinite separation, converting thepositive electric potential energy into kinetic energy. The two unlike charges, however, attract one another –if their separation is to be increased, a positive work must be done. In fact, the minimum amount of workthat must be done to create an infinite separation between the charges is equal to the magnitude of theoriginal negative electric potential energy.4. An equipotential surface must always cross an electric field line at right angles. Therefore, the equipotentialsurface at point 1 is concave to the right, the equipotential surface at point 2 is vertical, and the equipotentialsurface at point 3 is concave to the left.6. The electric field is either zero on this surface, or it is nonzero and perpendicular to the surface. If the electricfield had a nonzero component parallel to the surface, the electric potential would decrease as one moved inthe direction of the parallel component.8. Not necessarily. The electric potential energy of the two charges would be the same only if the charges arealso equal. In general, the electric potential energy is the charge times the electric potential. Therefore, justhaving equal potentials does not guarantee equal potential energy.10. Capacitors store electrical energy in the form of an electric field between their plates. This stored energy canremain in a capacitor even when a device such as a television is turned off. Accidentally touching theterminals can cause the sudden and potentially dangerous release of this energy.12. A capacitor stores charge of opposite sign in two different locations—though, of course, the net charge iszero. We can think of a capacitor, then, as storing a “charge separation,” along with the energy required tocause the charge separation in the first place.Solutions to Problems and Conceptual Exercises1. Picture the Problem: An electron is accelerated from rest in a region of space by a nonzero electric field.Strategy: Consider the connection between the electric field and the electric potential when answering the question.Solution: The electron moves in a direction opposite to the electric field. Because the electric field points in the direction of decreasing electric potential, it follows that the electron experiences an increasing potential.Insight: From an energy standpoint, electrical potential energy of the electron is converted into kinetic energy. This would correspond with a decrease in electric potential for a positive charge, but for the electron the work done by the field is W =− −e( )ΔV =eΔV =ΔK (equations 20-2 and 7-7) so that an increase in K corresponds to an increase in V.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.20 – 1Chapter 20: Electric Potential and Electric Potential Energy James S. Walker, Physics, 4th Edition2. Picture the Problem: A 4.5 µC charge moves in a uniform electric field rE = 4.1×105 N/C( )ˆx.Strategy: The change in electric potential energy of a charge that moves against an electric field is given by equation 20-1, ΔU =q0Ed. If the charge moves in the same direction as the field, the work done by the field is positive and ΔU =−W =−Fd =−q0Ed. If the charge moves in a direction perpendicular to the field, the work and change in potential energy are both zero. The change in potential energy from point A to point B is independent of the path the charge takes between those two points. Solution: 1. (a) The charge moves perpendicular to the field: ΔU = 02. (b) The charge moves in the same direction as the field: ΔU =−q0Ed =− 4.5×10−6C( )4.1×105N/C( )6.0m( )= −11 J3. (c) The potential energy changes by an amount equal to the sum of paths 1 and 2 above: ΔU =0−11 J = −11 JInsight: Note that U changes only along the direction parallel to the field. This is consistent with statement, “The electric field points downhill on the potential surface,” as described by figure 20-3 and the accompanying text.3. Picture the Problem: A uniform electric field rE = 6.8×105 N/C( )ˆx creates a change in the electric potential.Strategy: The change in electric potential is given by equation 20-4, ΔV =−EΔs , where Δs is equal to Δx because the field points in the ˆxdirection. There is no change in the electric potential along the ˆy and ˆzdirections. The change in electric potential from point A to point B is independent of the path taken between those two points. Solution: 1. (a) Here Δx =0because the path is in the ˆydirection: ΔV = 02. (b) Solve equation 20-4 for ΔV : ΔV =−E Δx =− 6.8×105N/C( )6.0m( )= −4.1×106 V3. (c) The potential changes by an amount equal to the sum of paths 1 and 2 above: ΔV =0−4.1×106 V = −4.1×106 V =−4.1 MVInsight: Note that the potential changes only along the direction parallel to the field. This is consistent with statement, “The electric field points downhill on the potential surface,” as described by figure 20-3 and the accompanying text.4. Picture the Problem: The cell membrane acts like a parallel plate capacitor, which is modeled in the diagram shown at right.Strategy: The field across the cell membrane is uniform, like that between the plates of a parallel plate capacitor. Use equation 20-4 to find the magnitude of the field that corresponds to the potential difference between the plates. Solution: 1. Apply equation 20-4 directly: E =ΔVΔs=ΔVd=0.070 V0.10×10−6 m= 7.0×105 V m2. The electric field points from higher to lower potential, so the electric field is directed inward across the cell membrane.Insight: The cell membrane was also modeled as a parallel plate capacitor in problems 52, 68, and 74 of Chapter 19.Copyright © 2010 Pearson Education, Inc. All rights reserved. This material is protected under all copyright laws as they currently exist. Noportion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher.20 – 2Chapter 20: Electric Potential and Electric Potential Energy James S. Walker, Physics, 4th Edition5. Picture the Problem: The two
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