30 1 Solve The potential difference V between two points in space is V V x f V x i E dx x x f x i where x is the position along a line from point i to point f When the electric field is uniform V E x x 1000 V m 0 30 m 0 10 m 200 V x f E dx x i x 30 2 Visualize Solve The potential difference V between two points on the y axis is V E dy y y f y i When the electric field is uniform the above result simplifies to In the present problem V E y y The y component of the electric field is y yf yi 0 05 m 0 05 m 0 10 m yE 50 000 V m V 50 000 V m 0 10 m 5 0 kV Assess Vf Vi 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 Ex 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 V 200 V m 2 0 m 1 0 m 200 V 3 0 m 2 0 m 300 V 1 2 Assess The potential difference is negative since the electric field points in the direction of decreasing potential 30 4 Model The potential difference is the negative of the area under the Ex vs x curve Visualize Please refer to Figure EX30 4 Solve The potential difference between the origin and x 3 0 m is V V x 3 0 m V x 0 m 100 V 1 0 m 0 m 200 V 3 0 m 1 0 m 1 2 1 2 150 V V Thus V 3 0 m 0 m 150 V 50 V 150 V 200 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 1 0 10 C 1 5 V 1 5 10 J U q V q V V i W 6 6 f Assess The work done by the escalator on the charge is stored as electric potential energy of the charge 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 W U q V q 6 1 0 10 V 0 V 1 60 10 19 6 C 1 0 10 V 1 6 10 13 J 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 W chem q 0 60 J 0 050 C 12 V 30 8 Solve The work done is the potential energy gained by the electron W U q V e V V 2 4 10 1 6 10 19 19 J C 1 5 V 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 Visualize Please refer to Figure EX30 9 The three equipotential surfaces correspond to potentials of 0 V 100 V and 200 V Solve The electric field component along a direction of constant potential is electric field component perpendicular to the equipotential surface is But the 0 V m dV ds E s cid 71 E V s 100 V 0 01 m 10 kV m The direction of the electric field vector is downhill perpendicular to the equipotential surfaces and directed to the left That is the electric field is 10 kV m to the left 30 10 Model The electric field is perpendicular to the equipotential lines and points downhill Visualize Please refer to Figure EX30 10 Three equipotential surfaces at potentials of 200 V 0 V and 200 V are shown Solve The electric field component perpendicular to the equipotential surface is The electric field vector is in the third quadrant 45 below the negative x axis That is E 20 kV m V s 200 V 0 01 m cid 71 E 20 kV m 45 below axis x 30 11 Model The electric field is the negative of the slope of the graph of the potential function Visualize Please refer to Figure EX30 11 Solve There are three regions of different slope For 0 cm x 10 cm and 20 cm x 30 cm V x 0 V m Ex 0 V m For 10 cm x 20 cm V x 100 V 100 V 0 20 m 0 10 m 2000 V m Ex 2000 V m dV ds the electric field is zero where the potential is not changing Assess Because E s 30 12 Model The electric field is the negative of the slope of the graph of the potential function Visualize Please refer to Figure EX30 12 Solve There are three regions of different slope For 0 cm x 1 cm For 1 cm x 2 cm For 2 cm x 3 cm V x 50 V 0 V 0 01 m 0 m 5000 V m Ex 5000 V m V dx 50 V 50 V 0 02 m 0 01 m 10 000 V m Ex 10 000 V m V x 50 V 0 V 0 03 m 0 02 m 5000 V m Ex 5000 V m and Ex are the negative of each other V x Assess 30 13 Visualize Solve The electric potential difference V between two points in a uniform electric field is V x f V x i E dx x E x x f x i Choosing xi 1 0 m and xf 1 0 m 1000 V 1000 V Ex 1 0 m 1 0 m Ex 1 0 kV m Alternatively xi 1 0 m and xf 1 0 m For this choice 1000 V 1000 V Ex 1 0 m 1 0 m Ex 1 0 kV m Assess The choice of initial and final positions does not change the physical nature of the electric field or the potential difference 30 14 Model The electric field is the negative of the derivative of the potential function E Solve s given potential a From Equation 30 11 the component of the electric field in the s direction is dV ds For the dV dx d dx x 100 V 200 x 2 E x x 200 V m V m At x …
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