Please do 1 and 2 1. 1/2 points |Previous Anawars HRW10 22.P003.ssm. My No The n
ID: 1871238 • Letter: P
Question
Please do 1 and 2
1. 1/2 points |Previous Anawars HRW10 22.P003.ssm. My No The nucleus of a lead-208 atom contains 82 protons. Assume that the nucleus is a sphere with radius 6.33 fm and with the charge of the protons uniformly spread through the sphere. At the nucleus surface, what are the magnitude and direction of the electric field produced by the protons? magnitude -3.20E21XNC - radialy outward The field of a positively charged particle points away from that particle. For a point on (or outside) the surface of a sphere of uniform charge, you can put all the charge at the center point. Additional Materials section 222 2. 0/1 points 1 Previous Anawars HRW10 22.P004 My No 3.00 x 10-7 C at x-6.00 cm, and particle 2 of charge +3.00 x 10-7 C at x-20.0 cm. Midway between the particles, what is their net electric field? Express your answer in Two particles are fixed to an x axis: particle 1 of charge vector form.) Did you draw the field vectors at the point, with the tails of the vectors anchored on the point? Did you draw the directions correctly? The field of a positively charged particle points away from that particle; the field of a negatively charged particle points toward that particle. Did you plug into the equation for the field of a point charge? Additional Materials Section 22.2Explanation / Answer
1) Number of protins=82
Charge on each proton=1.6*10-19 C
Total charge=82*1.6*10-19=131.2*10-19 C
Radius of sphere=6.33 fm=6.33*10-15 m
Elkectric field at surface = (9*109*131.2*10-19)/(6.33*10-15)2 = 2.95*1021 N/C
2) Charge at x=6cm is -3*10-7 C
Charge at x=20 cm is 3*10-7 C
x coordinate of midway point=(6+20)/2=13 cm
electric field at that point due to -3*10-7 C charge = 9*109*3*10-7/0.072=551020.4 N/C (along negative x axis)
electric field at that point due to 3*10-7 C charge = 9*109*3*10-7/0.072=551020.4 N/C (along negative x axis)
Total electric fiels at the point midway between two charges = 551020.4+551020.4=1102040.8 N/C (along negative x direction) = -1102040.8 i (in vector notation)