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Chapter 30

# Chapter 30 - 30.1 Solve The potential difference V between...

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30.1. Solve: The potential difference Δ V between two points in space is ( ) ( ) f i f i x x x V V x V x E dx Δ = = − where x is the position along a line from point i to point f. When the electric field is uniform, ( )( ) f i 1000 V/m 0.30 m 0.10 m 200 V x x x x V E dx E x Δ = − = − Δ = − = −

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30.2. Visualize: Solve: The potential difference Δ V between two points on the y -axis is f i y y y V E dy Δ = − When the electric field is uniform, the above result simplifies to . y V E y Δ = − Δ In the present problem, Δ y = y f – y i = 0.05 m – ( 0.05 m) = 0.10 m The y -component of the electric field is 50,000 V/m y E = − Δ V = – (–50,000 V/m)(0.10 m) = + 5.0 kV Assess: V f – V i = 5.0 kV shows that the potential at point f is higher than at point i. This is because the electric field is directed from the point at the higher potential to the point at the lower potential.
30.3. Model: The potential difference is the negative of the area under the E x vs x curve. Visualize: Please refer to Figure EX30.3. Solve: The potential difference between x = 1.0 m and x = 3.0 m is ( )( ) ( )( ) 1 200 V/m 2.0 m 1.0 m 200 V 3.0 m 2.0 m 300 V. 2 V Δ = − + = − Assess: The potential difference is negative since the electric field points in the direction of decreasing potential.

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30.4. Model: The potential difference is the negative of the area under the E x vs x curve. Visualize: Please refer to Figure EX30.4. Solve: The potential difference between the origin and x = 3.0 m is ( ) ( ) ( )( ) ( )( ) 1 1 3.0 m 0 m 100 V 1.0 m 0 m 200 V 3.0 m 1.0 m 2 2 150 V V V x V x Δ = = = = − + = − Thus ( ) ( ) 3.0 m 0 m 150 V 50 V 150 V 200 V V V = = − = − Assess: The potential decreases from the origin to x = 3.0 m.
30.5. Solve: The work done is exactly equal to the increase in the potential energy of the charge. That is, ( ) ( ) ( ) 6 6 f i 1.0 10 C 1.5 V 1.5 10 J W U q V q V V = Δ = Δ = = × = × Assess: The work done by the escalator on the charge is stored as electric potential energy of the charge.

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30.6. Solve: The Van de Graaff generator or the motor that runs the belt does work in lifting a positive ion ( q = e ) against the downward force on the positive charge that is moving up the belt. The work done is ( ) ( )( ) 6 19 6 13 1.0 10 V 0 V 1.60 10 C 1.0 10 V 1.6 10 J W U q V q = Δ = Δ = × = × × = × Assess: The work done by the generator in lifting the charge is stored as electric potential energy of the charge.
30.7. Solve: The emf is defined as the work done per unit charge by the charge escalator or the battery. That is, chem 0.60 J 12 V 0.050 C W q ε = = =

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30.8. Solve: The work done is the potential energy gained by the electron. ( ) ( ) 19 19 2.4 10 J 1.5 V 1.6 10 C W U q V e V V = = Δ = − Δ × ⇒ Δ = = − × The potential difference from the positive to the negative terminal is –1.5 V, so the emf of the solar cell is 1.5 V.
30.9. Model: The electric field points in the direction of decreasing potential and is perpendicular to the equipotential lines.

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