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 fi x x x VVx Vx E d x 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 xx x VE d xE x = −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 VE d y Δ= When the electric field is uniform, the above result simplifies to . y 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 () ( ) ( ) ( ) 11 3.0 m 0 m 100 V 1.0 m 0 m 200 V 3.0 m 1.0 m 22 150 V VVx Vx ⎛⎞ Δ= = = = − − + ⎜⎟ ⎝⎠ =− Thus ( ) 3.0 m 0 m 150 V 50 V 150 V 200 V VV = = 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, () ( ) 66 fi 1.0 10 C 1.5 V 1.5 10 J WU 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 ( ) ( )( ) 61 9 6 1 3 1.0 10 V 0 V 1.60 10 C 1.0 10 V 1.6 10 J WU 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 WUqV eV 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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Chapter 30 - 30.1 Solve The potential difference V between...

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