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chapt24 - Chapter 24 Electric Potential Problem 2 The...

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Chapter 24: Electric Potential Problem 2 The electric potential difference between the ground and a cloud in a particular thunderstorm is 1.2 × 10 9 V. What is the magnitude of the change in the electric potential energy (in multiples of the electron-volt) of an electron that moves between the ground and the cloud? Answer 1.2 × 10 9 eV. Solution Potential difference is work or energy per unit charge Δ V = Δ U/q or Δ U = q Δ V = (1 e)*(1.2 × 10 9 V) = 1.2 × 10 9 eV In SI units this is Δ U = 1.92 × 10 -10 J. Problem 4 When an electron moves from A to B along an electric field line in Fig. 24-29, the electric field does 3.94 × 10 -19 J of work on it. What are the electric potential differences (a) V B - V A , (b) V C - V A , and (c) V C - V B ? Answer (a) 2.46 V. (b) 2.46 V. (c) zero. Electric field lines A B C Equipotentials Figure 24-29 Solution (a) V B - V A = -W/q = - (3.94 × 10 -19 J)/(-1.6 × 10 -19 C) = 2.46 V. (b) Since C is on the same equipotential as B V C - V A = V B - V A = 2.46 V. (c) Since C and B are on the same equipotential V C - V B = 0.
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24-2 Chapter 24: Electric Potential Problem 5 An infinite nonconducting sheet has a surface charge density σ = 0.10 μ C/m 2 on one side. How far apart are the equipotential surfaces whose potentials differ by 50 V? Answer 8.8 mm. Solution Since the electric field due to an infinite sheet of charge is uniform the difference in potential between two points separated by a displacement Δ s is Δ V = - E ⋅Δ s If Δ s is parallel to E then Δ s = | Δ V|/E where E is the electric field due to an infinite sheet of charge: E = σ /2 ε 0 . Therefore Δ s = 2 ε 0 | Δ V|/ σ = 2*(8.85 × 10 -12 C 2 /N m 2 )*(50 V)/(0.10 μ C/m 2 ) Δ s = 8.8 mm. Problem 8 A graph of the x component of the electric field as a function of x in a region of space is shown in Fig. 24-30. The y and z components of the electric field are zero in this region. If the electric potential at the origin is 10 V, (a) what is the electric potential at x = 2.0 m, (b) what is the greatest positive value of the electric potential for points on the x axis for which 0 x 6.0 m, and (c) for what value of x is the electric potential zero? x (m) 20 0 -20 1 2 3 4 5 6 Figure 24-30 Answer (a) 30 V; (b) 40 V; (c) 5.5 m
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Chapter 24: Potential 24-3 Solution (a) The potential at x = 2 m is V (2 m ) = V (0) E x 0 2 m dx The integral is just the “area under the curve” from x = 0 to x = 2m. This “area” is equal to –20 volt. Therefore V(2 m) = 10 V – (-20 V) = 30 V. (b) The potential will continue to increase until x = 3 m after which the “area” becomes positive and the potential starts decreasing. Adding the area under the curve from 0 to 2 m to the potential at x = 0 gives V(3 m) = 10 V - (-30 V) = 40 V. (c) Starting from x = 3 m the potential decreases. The area under the curve between x = 3 m and x = 5.5 m is equal to 40 V. Therefore the potential will be zero at x = 5.5 m. Problem 9 The electric field in a region of space has the components E y = E z = 0 and E x = (4.00 N/C)x. Point A is on the y-axis at y = 3.00 m, and point B is on the x-axis at x = 4.00 m. What is the potential difference V B – V A ?
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