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Unformatted text preview: 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 200 V 20 kV/m 0.01 m V E s =  =  =  The electric field vector is in the third quadrant, 45 below the negative xaxis. That is, ( 29 20 kV/m, 45 belowaxis E x =  r 30.27. Model: Assume the battery is ideal. Visualize: Please refer to Figure EX30.27. Solve: For an ideal battery, the potential difference across the capacitor is the same as the emf of the battery. Thus, C Q V C = = E Q = C E = (10 10 6 F)(1.5 V) = 15.0 10 6 C 30.36. Visualize: Please refer to Figure P30.36. Solve: (a) The electric field points downhill. So, point A is at a higher potential than point B. (b) In a region that has a uniform electric field, Equation 30.3 gives the potential difference between two points: ( 29 f i x x V E dx E x x =  =  V B V A = (1000 V/m)(0.07 m) = 70 V That is, the potential at point A is 70 V higher than the potential at point B. 30.37. Solve: (a) (b) Equation 30.3 gives the potential difference between two points in space: ( 29 ( 29 ( 29 f i 30 cm 20 cm 0.30 m 2 2 2 0.20 m 1000 V/m 1000 1000 V 0.30 m 0.20 m V 65 V 2 2 x x x V E dx x dx x =  =  = + = +  = + Assess: E is positive for negative x but negative for positive x , so the potential difference depends on the square of the positions. 30.38. Solve: (a) (b) Equation 30.3 gives the potential difference between two points in space: ( 29 ( 29 ( 29 ( 29 f f f i i i 2 2 2 f i f i 5000 V/m 5000 V 2500 V 2 x x x x x x x x V V x V x E dx x dx...
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 Spring '12
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