646_PartUniversity Physics Solution

646_PartUniversity Physics Solution - Electric Charge and...

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Electric Charge and Electric Field 21-39 The charge is negative, so the field at the origin is directed toward the small segment. The small segment is located at angle θ as shown in the sketch. The electric field due to dQ is shown in Figure 21.97b, along with its components. 2 0 1 4 dQ dE a π = P 22 0 2 Q dE d a = P Figure 21.97b ( ) 0 cos / 2 cos x dE dE Q a d πθ == P () /2 0 0 00 0 cos sin 2 xx QQ Q Ed E d aa a θθ ππ = = ∫∫ PP P ( ) 0 sin sin y dE dE Q a d P 0 0 0 sin cos 2 yy Q E d a = = P EVALUATE: Note that , x y E E = as expected from symmetry. 21.98. IDENTIFY: Apply 0 x F = and 0 y F = to the sphere, with x horizontal and y vertical. SET UP: The free-body diagram for the sphere is given in Figure 21.98. The electric field E G of the sheet is directed away from the sheet and has magnitude 0 2 E σ = P (Eq.21.12). EXECUTE: 0 y F = gives cos Tm g α = and cos mg T = . 0 x F = gives 0 sin 2 q T = P and 0 2s i n q T = P . Combining these two equations we have 0 cos 2 sin mg q = P and 0 tan 2 q mg = P . Therefore, 0 arctan 2 q mg ⎛⎞ = ⎜⎟ ⎝⎠ 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 P due to a finite wire. These fields add by vector addition. SET UP: Each field has magnitude 0 1 4 Q x xa + 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 E net = 2 E 1 cos 45° = 0 1 2c o s 4 5 4 Q xx a ° + P . In part (b), the electrical force on an electron at P is eE . EXECUTE: (a) The net field is E net = 0 1 o s 4 5 4 Q ° + P . E net = ( )( ) 92 2 6 2 9.00 10 N m /C 2.50 10 C cos45 (0.600 m) (0.600 m) (0.600 m) ×⋅ × ° + = 6.25 × 10 4 N/C. The direction is 225° counterclockwise from an axis pointing to the right through the positive wire.
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21-40 Chapter 21 (b) F = eE = (1.60 × 10 -19 C)(6.25 × 10 4 N/C) = 1.00 × 10 –14 N, opposite to the direction of the electric field, since the electron has negative charge. EVALUATE: Since the electric fields due to the two wires have equal magnitudes and are perpendicular to each other, we only have to calculate one of them in the solution. 21.100. IDENTIFY: Each sheet produces an electric field that is independent of the distance from the sheet. The net field is the vector sum of the two fields. SET UP: The formula for each field is 0 /2 , E σ = P and the net field is the vector sum of these, net 00 0 22 2 B ABA E σσ ± =±= PP P , where we use the + or – sign depending on whether the fields are in the same or opposite directions and B and A are the magnitudes of the surface charges. EXECUTE: (a) The two fields oppose and the field of B is stronger than that of A , so E net = 0 2 B AB A −= P = () 12 2 2 11.6 C/m 9.50 C/m 28 .85 10 C/Nm μμ ×⋅ = 1.19 × 10 5 N/C, to the right.
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646_PartUniversity Physics Solution - Electric Charge and...

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