charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4.00 × 10 4 N/C. What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 0.670 m upward (c) 2.60 m at an angle of 45.0° downward from the horizontal? Textbook Question Chapter 18, Problem 1P A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4.00 × 104 N/C. What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 0.670 m upward (c) 2.60 m at an angle of 45.0° downward from the horizontal?
A charge of 28.0 nC is placed in a uniform electric field that is directed vertically upward and that has a magnitude of 4.00 × 104 N/C. What work is done by the electric force when the charge moves (a) 0.450 m to the right; (b) 0.670 m upward (c) 2.60 m at an angle of 45.0° downward from the horizontal?
(a)
0.450m0.450 m
0.450m0.450 mzero⎯⎯⎯⎯⎯⎯⎯zero_
28.0nC28.0 nC4.00×104N/C4.00×104 N/C
W=qEscosϕW=qEscosϕ
WWqqEEssϕϕqEqE
ϕ=90°ϕ=90°cos90°=0cos90°=0
00cosϕcosϕ
W=0W=0
0.450m0.450 mzero⎯⎯⎯⎯⎯⎯⎯zero_
(b)
0.670m0.670 m
0.670m0.670 m7.50×10−4J⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯7.50×10−4 J_
28.0nC28.0 nC4.00×104N/C4.00×104 N/C0.670m0.670 m
W=qEscosϕW=qEscosϕ
ϕ=0°ϕ=0°
28.0nC28.0 nCqq4.00×104N/C4.00×104 N/CEE0.670m0.670 mss0°0°ϕϕWW
W=(28.0nC)(4.00×104N/C)(0.670m)(cos0°)=(28.0nC×1C109nC)(4.00×104N/C)(0.670m)(cos0°)=7.50×10−4JW=(28.0 nC)(4.00×104 N/C)(0.670 m)(cos0°)=(28.0 nC×1 C109 nC)(4.00×104 N/C)(0.670 m)(cos0°)=7.50×10−4 J
0.670m0.670 m7.50×10−4J⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯7.50×10−4 J_
(c)
2.60m2.60 m45.0°45.0°
2.60m2.60 m45.0°45.0°−2.06×10−3J⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯−2.06×10−3 J_
28.0nC28.0 nC4.00×104N/C4.00×104 N/C2.60m2.60 m45.0°45.0°
W=qEscosϕW=qEscosϕ
45.0°45.0°90.0°+45.0°=135.0°90.0°+45.0°=135.0°
28.0nC28.0 nCqq4.00×104N/C4.00×104 N/CEE2.60m2.60 mss135.0°135.0°ϕϕWW
W=(28.0nC)(4.00×104N/C)(2.60m)(cos135.0°)=(28.0nC×1C109nC)(4.00×104N/C)(2.60m)(cos135.0°)=−2.06×10−3JW=(28.0 nC)(4.00×104 N/C)(2.60 m)(cos135.0°)=(28.0 nC×1 C109 nC)(4.00×104 N/C)(2.60 m)(cos135.0°)=−2.06×10−3 J
2.60m2.60 m45.0°45.0°−2.06×10−3J⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯−2.06×10−3 J_
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