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(a) Figure 24-10a shows two points i and fin a uniform electric field overset(to)(E ). The points lie on the same electric field line (not shown) and are separated by a distanced. Find the potential difference V_(f)- V_(i) by moving a positive test charge q_(0) from i to f along the path shown, which is parallel to the field direction. (b) Now find the potential difference V_(f)- V_(i) by moving the positive test charge q_0 from i to f along the path icf shown in Fig. 24-10b. |
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Answer» Solution :We can find the potential difference between any two points in an electric field by INTEGRATING `overset(to)( E) .d overset(to) (s)` along a path connecting those two points ACCORDING to Eq. 24-20. Calculations: We have actually already DONE the calculation for such a path in the direction of an electric field line in a uniform field when we derived Eq. 24-23. With slight changes in NOTATION, Eq. 24-23 gives us `V_(f)- V_(i)= - Ed.`(Answer) Calculations: The Key Idea of (a) applies here too, except now we move the test charge along a path that consists of two lines: is and cf. .At all points along line is, the displacement `d overset(to)s` of the test charge is perpendicular to `overset(to) (E)`. Thus, the angle `theta` between `overset(to)(E) and d overset(to)(s)` is 90°, and the dot product `overset(to)(E). d overset(to)(s)` is 0. Equation 24-20 then tells us that points i and c are at the same potential: `V_(c ) - V_(i)= 0`. Ah, we should have seen this COMING. The points are on the same equipotential surface, which is perpendicular to the electric field lines. For lines `cf` we have `theta=45^@` and, from Eq `24-20`, `V_(f)- V_(i)-int_(c)^(f) overset(to)( E).doverset(to)(s)=-int_(c)^(f) E ( cos 45^@) ds` `=- E ( cos 45^@) int_(c )^(f) ds`. |
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