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Electric expressions9/2/2023 ![]() It follows that an electron accelerated through 50 V gains 50 eV. Īn electron accelerated through a potential difference of 1 V is given an energy of 1 eV. (Note that in terms of energy, “downhill” for the electron is “uphill” for a positive charge.) Since energy is related to voltage by Δ U = q Δ V Δ U = q Δ V, we can think of the joule as a coulomb-volt.ġ eV = ( 1.60 × 10 −19 C ) ( 1 V ) = ( 1.60 × 10 −19 C ) ( 1 J/C ) = 1.60 × 10 −19 J. The electron gains kinetic energy that is later converted into another form-light in the television tube, for example. An electron is accelerated between two charged metal plates, as it might be in an old-model television tube or oscilloscope. It is useful to have an energy unit related to submicroscopic effects.įigure 7.13 shows a situation related to the definition of such an energy unit. ![]() The particle may do its damage by direct collision, or it may create harmful X-rays, which can also inflict damage. For example, even a tiny fraction of a joule can be great enough for these particles to destroy organic molecules and harm living tissue. But on a submicroscopic scale, such energy per particle (electron, proton, or ion) can be of great importance. The energy per electron is very small in macroscopic situations like that in the previous example-a tiny fraction of a joule. How many electrons would go through a 24.0-W lamp each second from a 12-volt car battery? The Electron-Volt To have a physical quantity that is independent of test charge, we define electric potential V (or simply potential, since electric is understood) to be the potential energy per unit charge: But we do know that because F → = q E → F → = q E →, the work, and hence Δ U, Δ U, is proportional to the test charge q. d → and the direction and magnitude of F → F → can be complex for multiple charges, for odd-shaped objects, and along arbitrary paths.Calculating the work directly may be difficult, since W = F → Therefore, although potential energy is perfectly adequate in a gravitational system, it is convenient to define a quantity that allows us to calculate the work on a charge independent of the magnitude of the charge. (The default assumption in the absence of other information is that the test charge is positive.) We briefly defined a field for gravity, but gravity is always attractive, whereas the electric force can be either attractive or repulsive. Recall that earlier we defined electric field to be a quantity independent of the test charge in a given system, which would nonetheless allow us to calculate the force that would result on an arbitrary test charge. Apply conservation of energy to electric systems.Describe systems in which the electron-volt is a useful unit.Calculate electric potential and potential difference from potential energy and electric field. ![]()
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