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# Page 10 - 13 The graphof an electric ﬁeld in one...

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Unformatted text preview: 13.) The graphof an electric ﬁeld in one dimension is shown to the right. What would the corresponding electrical potential graph look like? ﬁglntign: The electrical potential difference between any two points is the area under the electric ﬁeld versus position graph (AV: E-dr). Assuming the electrical potential is some positive value, it will get bigger between zero and a, will get still bigger but not as fast between a and 2a, will not change at all between 2:; and 3a, then will get bigger again, etc. The graph is shown to the right. 14.) The graph of an electrical potential ﬁeld in one dimension is shown to the right. What would the corresponding electric ﬁeld graph look like? Solution: The slope of the electrical potential function at any point is equal to the electric field function (E = -dV/dx). That slope is constant and negative in the section from zero to 2:1, then is zero, then constant and positive from 3a to 4a., then varies from positive to zero to negative in a relatively linear way. The graph is shown to the right. 15.) A charge q in a constant electric ﬁeld sits at a point where the electrical potential is V1, where V1 is positive. It accelerates from rest to a position at which the electrical potential is 2V1. At that point, the particle's velocity is v. a.) Is the charge positive or negative? Shim: The charge is negative. How so? It accelerates from a lower potential (i.e. V ) to higher potential (i.e. 2V ). Negative charges do that—-positive 1 1 charges accelerate from higher to lower electrical potentials. b.) If (1 had gone to a position where the electrical potential was 3V1, how fast would it have been going (do this in terms of the known velocity v)? §glutionz This is a conservation of energy problem. In the original problem, the electrical potential difference was 2V1 - VI = V]. That means the work done was W = -AV = -(vq)V . Also, given that the initial velocity was zero, the kinetic I l g____ﬂ4 ...
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