Electric Charge and Electric Field21-39The charge is negative, so the field at the origin is directed toward the small segment. The small segment is located at angle θas shown in the sketch. The electric field due to dQis shown in Figure 21.97b, along with its components. 2014dQdEaπ=P2202QdEdaθπ=PFigure 21.97b ()220cos/2cosxdEdEQadθπθ θ==P()/2/ 202222220000cossin222xxQQQEdEdaaaππθ θθπππ====∫∫PPP()220sin/2sinydEdEQadθπθ θ==P()/2/ 202222220000sincos222yyQQQEdEdaaaππθ θθπππ===−=∫∫PPPEVALUATE:Note that ,xyEE=as expected from symmetry. 21.98. IDENTIFY:Apply 0xF=∑and 0yF=∑to the sphere, with xhorizontal and yvertical. SET UP:The free-body diagram for the sphere is given in Figure 21.98. The electric field EGof the sheet is directed away from the sheet and has magnitude 02Eσ=P(Eq.21.12). EXECUTE:0yF=∑givescosTmgα=andcosmgTα=. 0xF=∑gives 0sin2qTσα=Pand 02sinqTσα=P. Combining these two equations we have 0cos2sinmgqσαα=Pand 0tan2qmgσα=P. Therefore, 0arctan2qmgσα⎛⎞=⎜⎟⎝⎠P. EVALUATE:The electric field of the sheet, and hence the force it exerts on the sphere, is independent of the distance of the sphere from the sheet. Figure 21.98 21.99. IDENTIFY:Each wire produces an electric field at Pdue to a finite wire. These fields add by vector addition. SET UP:Each field has magnitude 22014Qxxaπ+P. The field due to the negative wire points to the left, while the field due to the positive wire points downward, making the two fields perpendicular to each other and of equal magnitude. The net field is the vector sum of these two, which is Enet= 2E1cos 45° = 22012cos454Qxxaπ°+P. In part (b), the electrical force on an electron at Pis eE. EXECUTE:(a)The net field is Enet= 22012cos454Qxxaπ°+P. Enet= ()()9226222 9.0010 N m /C2.5010C cos45(0.600 m)(0.600 m)(0.600 m)−×⋅×°+= 6.25 ×104N/C. The direction is 225° counterclockwise from an axis pointing to the right through the positive wire.
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