PHYS-102 Homework-03 Solutions P20.1(a) Energy of the proton-field system is conserved as the proton moves from high to low potential, which can be defined for this problem as moving from 120 V down to 0 V. Ki+Ui+ΔEmech=Kf+Uf0+qV+0=12mvp2+0 1.60×10−19C()120 V1 J1 V⋅C⎛⎝⎜⎞⎠⎟=121.67×10−27kgvp2vp=1.52×105ms(b) The electron will gain speed in moving the other way, from Vi=0 to Vf=120 V : Ki+UiEmech=Kf+Uf0+0+0=12mve2+qV0=129.11×10−31kgve2+−1.60×10−19C120 J Cve=6.49×106P20.12(a) V=keq1r1+keq2r2=2keqr⎛⎝⎜⎞⎠⎟V=28.99×109N⋅m2C22.00×10−6C()1.00 m2+0.500 m2⎛⎝⎜⎜⎞
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This note was uploaded on 09/03/2008 for the course PHYS 102 taught by Professor N/a during the Spring '08 term at Drexel.