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In this example we will calculate the potential in a region of space where the electric field is uniform. In particular, we will find the potential at any height y between two charged parallel plates. igure V <1 of 1 SOLUTION SET UP (Figure 1) shows our sketch. As before, we point the y axis upward. The electric field is uniform and directed vertically downward. We choose the potential V to be zero at y = 0 (point b in our sketch). The potential increases linearly as we move toward the upper plate. SOLVE The potential energy U for a test charge q' at a distance y above the bottom plate is given by U= q'Ey. The potential V at point y is the potential energy per unit charge, V = U/q', so V = Ey. Even if we had chosen a different reference level (at which V = 0), it would still be true that Vy - V = Ey. At point a, where y = d and Vy= Va. Va Vs Ed and E = Va-V = Va REFLECT The magnitude of the electric field equals the potential difference between the plates divided by the distance between them. (Caution! This relationship holds only for the parallel-plate arrangement described here, in which the electric field is uniform.) Part A - Practice Problem: Suppose that Constants Lohosen eeen the potential to be ser aper plate where a d Derive on expression for the potential at opy value of a

In this example we will calculate the potential in a region of space where the electric field is uniform. In particular, we will find the potential at any height y between two charged parallel plates. igure V <1 of 1 SOLUTION SET UP (Figure 1) shows our sketch. As before, we point the y axis upward. The electric field is uniform and directed vertically downward. We choose the potential V to be zero at y = 0 (point b in our sketch). The potential increases linearly as we move toward the upper plate. SOLVE The potential energy U for a test charge q' at a distance y above the bottom plate is given by U= q'Ey. The potential V at point y is the potential energy per unit charge, V = U/q', so V = Ey. Even if we had chosen a different reference level (at which V = 0), it would still be true that Vy - V = Ey. At point a, where y = d and Vy= Va. Va Vs Ed and E = Va-V = Va REFLECT The magnitude of the electric field equals the potential difference between the plates divided by the distance between them. (Caution! This relationship holds only for the parallel-plate arrangement described here, in which the electric field is uniform.) Part A - Practice Problem: Suppose that Constants Lohosen eeen the potential to be ser aper plate where a d Derive on expression for the potential at opy value of a

College Physics
College Physics
11th Edition
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Raymond A. Serway, Chris Vuille
Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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In this example we will calculate the potential in a region of space where the electric field is uniform. In particular, we will find the potential at any height y between two charged parallel plates. Figure +X+ a 9' + bo }/-//-/-/-/A/ 1 of 1 SOLUTION SET UP (Figure 1) shows our sketch. As before, we point the y axis upward. The electric field is uniform and directed vertically downward. We choose the potential V to be zero at y = 0 (point b in our sketch). The potential increases linearly as we move toward the upper plate. SOLVE The potential energy U for a test charge q' at a distance y above the bottom plate is given by U = q'Ey. The potential V at point y is the potential energy per unit charge, V = U/q', so V = Ey. Even if we had chosen a different reference level (at which V = 0), it would still be true that Vy - V₂ = Ey. At point a, where y = d and Vy = Va, Va – Vi = Ed and Part A - Practice Problem: V REFLECT The magnitude of the electric field equals the potential difference between the plates divided by the distance between them. (Caution! This relationship holds only for the parallel-plate arrangement described here, in which the electric field is uniform.) Submit Suppose that we had chosen the potential to be zero at the upper plate, where y = d. Derive an expression for the potential at any value of y. xpress your answer in terms of variables E, d, and y. [ΨΕΙ ΑΣΦ Provide Feedback E Request Answer Constants Va-Vb d ? Vab d Next >
Transcribed Image Text:In this example we will calculate the potential in a region of space where the electric field is uniform. In particular, we will find the potential at any height y between two charged parallel plates. Figure +X+ a 9' + bo }/-//-/-/-/A/ 1 of 1 SOLUTION SET UP (Figure 1) shows our sketch. As before, we point the y axis upward. The electric field is uniform and directed vertically downward. We choose the potential V to be zero at y = 0 (point b in our sketch). The potential increases linearly as we move toward the upper plate. SOLVE The potential energy U for a test charge q' at a distance y above the bottom plate is given by U = q'Ey. The potential V at point y is the potential energy per unit charge, V = U/q', so V = Ey. Even if we had chosen a different reference level (at which V = 0), it would still be true that Vy - V₂ = Ey. At point a, where y = d and Vy = Va, Va – Vi = Ed and Part A - Practice Problem: V REFLECT The magnitude of the electric field equals the potential difference between the plates divided by the distance between them. (Caution! This relationship holds only for the parallel-plate arrangement described here, in which the electric field is uniform.) Submit Suppose that we had chosen the potential to be zero at the upper plate, where y = d. Derive an expression for the potential at any value of y. xpress your answer in terms of variables E, d, and y. [ΨΕΙ ΑΣΦ Provide Feedback E Request Answer Constants Va-Vb d ? Vab d Next >
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