Ch 30 Physics for Scientists and Engineers

# Ch 30 Physics for Scientists and Engineers - POTENTIAL AND...

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POTENTIAL AND FIELD 30.1. Solve: The potential difference AV between two points in space is 9 AV = V(xf) - V(x,) = -IE, dx x, where x is the position along a line from point i to point f. When the electric field is uniform, AV = -Exjd. =-E,& = -(lo00 V / m)(0.30 m - 0.10 m) = -200 V xr X, ii 30.2. Visualize: Solve: The potential difference AV between two points on the y-axis is = -7 E, dy x When the electric field is uniform, the above result simplifies to AV = -E, Ay. In the present problem, Ay = yr - yj = 0.05 m - (-0.05 m) = 0.10 m The ycomponent of the electric field is E, = -50,000 V / m 3 AV= - (-50,000 V/m)(O.lO m) = +5000 V Assess: field is directed from the point at the higher potential to the point at the lower potential. V, - = 5000 V shows that the potential at point f is higher than at point i. This is because the electric 30-1 . ~ .. _._

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30-2 Chapter 30 v= +loooV - x (m) 30.3. Visualize: -1Ooov Ex -1.0 0 + 1.0 f~--------------------------------i Solve: 30.4: The electric potential difference AV between two points in a uniform electric field is given by Equation V(xJ - V(XJ = -Ex (xf - x,) Choosing x, = -1 .O m and = +1 .O m, +lo00 V- (-1000 V)=-E,[l.O m-(-1.0 m)]a E,=-lOOO Vlrn Alternatively, x, = 1 .O m and = -1 m. For this choice, -1000 V - (+lo00 V) = -Ex [-1.0 m - (1.0 m)] * Ex= -1000 Vlrn Assess: The choice of initial and final positions does not change the physical nature of the electric field or the potential difference. 30.4. Solve: potential, Model: The electric field is the negative of the derivative of the potential function. From Equation 30.8, the component of the electric field in the s-direction is E, = -dV/ds. For the given dV du - d = -(loox' v) = 200x V m - E, = -200x Atx= Om, = 0 Vlm ,and at x= 1 m, E, =-200 (1) V/m=-200 Vlrn. Assess: The potential increases with x, so the electric field must point in the -x direction. 30.5. Solve: Model: The electric field is the negative of the slope of the graph of the potential function. There are three regions of different slope. For 0 cm < x < 10 cm and 20 cm < x < 30 cm, -=OVlm* E,=OVIm Ax For lOcm <x< ?Ocm, AV -100 V - (100 V) = -2000 V I m 3 = +2000 Vlrn _- - Ax 0.20m-0.lOm I 10 20 30 Assess: Because E, = -dV/ds , the electric field is zero where the potential is not changing. 30.6. Solve: Model: The electric field is the negative of the slope of the graph of the potential function. There are three regions of different slope. For 0 cm < x < 1 cm, -= 50v-0v =5000~/m ~E~=-~OOOV/~ Ax 0.01m - Om For1 cm<x<?cm,, = -10,000 V / m * = 10,000 Vlm AV -50 V - (50 V) dx 0.02 m-0.01 m For 2 cm <x < 3 cm, = 5000 VI m a E,= -5000 Vlm AV OV -(-5OV) 0.03 m - 0.02 m
Potential and Field 30-3 I Assess: AV/& and E, are the negative of each other. 30.7. Model: equipotential lines. Visualize: and 200 V. Solve: electric field component perpendicular to the equipotential surface is The electric field points in the direction of decreasing potential and is perpendicular to the Please refer to Figure Ex30.7. The three equipotential surfaces correspond to potentials of 0 V, 100 V, The electric field component along a direction of constant potential is = -dV/ds = 0 V / m. But, the 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 10,000 V/m to the left.

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## This homework help was uploaded on 04/07/2008 for the course PHSX 211/212 taught by Professor Medvedev during the Spring '08 term at Kansas.

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Ch 30 Physics for Scientists and Engineers - POTENTIAL AND...

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