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Most of us have seen a topographical map: Electric Potential, Energy, Capacitance 16-2 Equipotential Surfaces The contour lines show points on the mountain that are at the same elevation. The yellow contour line connects points that are 14,200 feet above sea level. Since my weight is 180 lbs, when I am standing on that contour line I have a potential energy of U G = mgh = (180)(14200) = 2,556,000 ftlbs FYI: If the altitude of Milwaukee is 600 ft, my potential energy in Milwaukee is U = (180)(600) = 108,000 ftlbs, making U = 2,448,000 ftlbs. Where did this energy come from? T or F: The gravitational field vector g is perpendicular to every point on the contour line.

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The 2D contour lines define a 3D surface: Topic 6.2 Extended E – Equipotential surfaces Here's the 3D view from the top of Mount Elbert:

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It is easier to visualize how these equipotential surfaces fit together to form a 3D image of the actual surface if we concentrate on the vicinity of the crest of Mount Elbert: Electric Potential, Energy, Capacitance 16-2 Equipotential Surfaces

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Let's look at the equipotential surfaces from 14,200 feet up to the top of Mount Elbert:

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Rotation, tilting, and stacking of these equipotential surfaces will produce what appears to be a 3D image of Mount Elbert: 14200 14240 14280 14320 14360 14400

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Rotation, tilting, and stacking of these equipotential surfaces will produce what appears to be a 3D image of Mount Elbert:

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Topic 6.2 Extended E – Equipotential surfaces Of course, on the planetary scale the equipotential surfaces will be spherical, not flat. And the contour lines will look like this: T or F: The gravitational field vector g is perpendicular to every point on the contour line. g g g Question: Why are the surfaces farther from the center farther apart?

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Topic 6.2 Extended E – Equipotential surfaces The negative point charge acts as a planet, setting up equipotential surfaces in the same way: And the contour lines about a negative point charge will look like this: T or F: The electric field vector E is perpendicular to every point on the equipotential surface. E E E Question: Why are the surfaces farther from the center farther apart?

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Topic 6.2 Extended E – Equipotential surfaces Suppose we have two parallel plates separated by 15 mm and charged by a 9 V battery. (a) How far apart would the equipotential surfaces be between the plates, if their potential difference was to be 0.10 V? Since the electric field is constant between the plates the equipotential surfaces will be evenly spaced (unlike those around a point charge). - - - - - - - - - - + + + + + + + + + + 0.015 m V = Ed first we find the value of the electric field... 9 = E(0.015) E = 600 V/m now we find the distance x between the surfaces... V = ExEx 0.10 = 600( x) x = 1.67 10 -4 m

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Topic 6.2 Extended E – Equipotential surfaces Suppose we have two parallel plates separated by 15 mm and charged by a 9 V battery. (b) If we assign the value of 0 V to the negative plate, where is the equipotential surface with a potential of +3.75 V located? We can use the formula -------------------- ++++++++++++++++++++ 0.015 m V = Ex:Ex: 3.75 = 600 x: x = 0.00625 m 0.00625 m

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Topic 6.2 Extended E – Equipotential surfaces -------------------- ++++++++++++++++++++ 0.015 m 0.00625 m If we know the potentials and the geometry of the equipotential surfaces surrounding a charged object we can find the value of the E-field using E = - VxVx Electric Field From Potential For example, suppose we are given that the voltage difference between the negative plate and the red equipotential surface is 3.75 V (it is) then the above formula gives us E = - VxVx = - V - V x - x 0 0 = - 3.75 - 0.00625 - 0 = -600 V/m this is the expected value FYI: The negative in the above formula gives the correct direction of E.

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Topic 6.2 Extended E – Equipotential surfaces Sketch in the equipotential surfaces between the two charges. + - FYI: If you are given the equipotential surfaces, you can construct the E- field lines.

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Topic 6.2 Extended E – Equipotential surfaces FYI: The definition of the eV uses an electron. But any particle that has a charge which is an integral number can use electron volts as an energy quantity. T HE E LECTRON V OLT eV Recall the relationship between voltage (electric potential) and potential energy: U = q V. If we place an charge in an electric field, it will accelerate, changing its kinetic energy. From energy considerations we have K + U = 0 K = - U K = -q V The Kinetic Energy Change of a Charge We define the electron volt (eV) as the kinetic energy gained by an electron accelerated through a potential difference of exactly 1 volt: 1 eV = -(-1.6 10 -19 C)(1.000 V) K = -q V 1 eV = 1.6 10 -19 J Electron Volt FYI: The electron volt is NOT an SI unit of energy, but it is used often for describing the energy of atomic-sized particles.

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