Question

thanks you so much for your help in advance, if you could do both that would be so helpful as i am out of questions, thank you ill be sure to rate

My Notes 4. -/1 points SERPSE 10 24.1.OP.005.MI. An electron moving parallel to the x axis has an initial speed of 4.14 x 10 m/s at the origin. Its speed is reduced to 1.50 x 10 m/s at the point x = 2.00 cm. (a) Calculate the electric potential difference between the origin and that point. Volts (b) Which point is at the higher potential? the origin the point 2.00 cm both have the same potential

Answer 1

An election moves parallel to n-axis initial speed u = 4.148 106 mis final speed v= 1.50 xios mis Similanly initial position 120 con - final position Vf = 2 cm o use work enengy theonem At AV-O -) Kf-Ki + AU=0 :.:) ģm V- miv? tavio on à m ( V? - V; 2) + LAV = 0 . .. mass of é m= 9.1x10 lug .. e charge qe = -1.6X1044 on - 9e av = à m (ve - V; 2) AVE * (v - V;2) Put values, we get SV = -4 9.1.3 ou [ 0.52083)?- (14.02.2018] vody = 2.844410"(22.5010"– 17. 14 xnovy v

-.on AV = 2.844 x10? x 1010 ( 22.5 - 1714) v = 2.844 x102 (-169115] V . -4810.6 3 x10? V ~ -upi iu ii Ava -48-1 Voets nj) ② Here the potential diffenence is negative. so the potential at onigin is greater than potential at point u= 2.00cm. Thus (at origin) potential is high. Canj ) (As ben Chegy guidelines, i can do only 1 eiene,