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# 22-5and6 - Nonuniformiy charged solid sphere Suppose the...

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Unformatted text preview: Nonuniformiy charged solid sphere. Suppose the charge density of the solid sphere in Fig. 22—12. Example 22—4, is given by pH = era, where or is a constant. [e] Find o in terms of the total charge Q on the sphere and its radius rﬂ. (It) Find the electric field as a function of r inside the sphere MW We divide the sphere up into concentric thin shells of thickness dr as shown in Fig. 22—14, and integrate {e} setting Q = fpﬁ oil” and {b} using lGrauss's law. SDLUTIDN {e} A thin shell of radius r and thickness dr {Fig 22—14} has volume oTlHr = 41er ctr. The total charge is given lay ’” 41TH! 5 Q = JpEdV = Inferrzlldlnrzdr] = 41rotJ r‘lrfr = 5 ru. 0 I] Thus at = sot-1mg. [h] To find E inside the sphere at distance r from its center, we apply Gauss’s law to an imaginary sphere of radius r which will enclose a charge .1- r z 2 r 5g. 2 2 r5 Qend = PE fill"r = [ﬂrl4srr {fr = Sr 4111-;- [fr 2 9—5.. a o "n g alert-n By symmetry. E will be the same at all points on the surface of a sphere of radius r, so Gauss’s law gives ¥E ' if; ancl Eu 2 r5 51(er — owl so E = er . 4wsﬂr3 Long uniform line of charge. A very long straight wire possesses a uniform positive charge per unit length. x. Calculate the electric field at points near (but outside} the wire~ far from the ends. APPROACH Because of the symmetry, we expect the field to be directed radially outward and to depend only on the perpendicular distance, R, from the wire. Because of the cylindrical symmetry, the field will he the same at all points on a gaussian surface that is a cylinder with the wire along its axis. Fig. 22—15. E is perpendicular to this surface at all points. For Gauss‘ s law we need a closed surface so we include the flat ends of the cylinder. Since E ts parallel to the ends, there is no flux through the ends [the cosine of the angle between E and out on the ends is cos EIIJ" = ﬂ}. SDlU'I'IDN For our chosen gaussian surface Ga uss‘s law gives Quins] Ill—E En En dis-ail = E{2nﬂf} = where f is the length of our chosen gaussian surface (f :3: length of wire]. and 21TH is its circumference. HeHCe 1 A 21111] R NOTE This is the same result we found in Example 21—11 using CoulomFs law [we used 16 there instead of R), but here it took much less effort. Again we see the great power of Gauss’s law.1 NOTE Recall from Chapter 10. Fig. ill-2, that we use R for the distance of a particle from an axis (cylindrical symmetry}, but lower case r for the distance from a point [usually the origin Ll]. E= ...
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