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10_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

# 10_Ch 16 College Physics ProblemCH16 Electrical Energy and Capacitance

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44 CHAPTER 16 From Newton’s second law, y y F mg qE qE ag mm m Σ −− == = +  . Equating this to the earlier result gives 0 2 y y v qE mt  =− + =  , so the electric field strength is ( ) 0 23 2 s 9.80 m s 1.95 10 N C y v m Eg qt  − = ×   () ( 6 220 .1 m 2.00 kg 5.00 10 C 4.10 s = × Thus, ) ( ) 34 max max 20.6 m 1.95 4.02 10 V 40.2 kV Vy E ∆= ∆ = × =× = 16.11 (a) ( )( ) 92 2 1 9 7 -2 8.99 10 N m C 1.60 10 C 1.44 10 V 1.00 10 m e kq r ×⋅ × = × × V (b) ( ) 21 2 1 2 1 9 8 11 8.99 C 1.60 10 C 0.020 0 m 0.010 0 m 7.19 10 V ee e VV V k q rr r r =−=−= =× ⋅ × × 16.12 12 e qq VV V k =+= + where 1 0.60 m
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