show all steps please Two 1.9-cm-diameter disk face each other, 2.2 mm apart. Th
ID: 1537372 • Letter: S
Question
show all steps please
Two 1.9-cm-diameter disk face each other, 2.2 mm apart. They are changed to plusminus 16 nC. What is the electric field strength at the midpoint between the centers of the disks? Express your answer to two significant figures and include the appropriate units. A proton is shot from the center of the negative disk toward the center of the positive disk. What launch speed must the proton have to just barely reach the positive disk? Express your answer to two significant figures and include the appropriate units.Explanation / Answer
the potential of the capacitor is V = E*d = 4.1*10^5*1.6*10^-3 V = 656 V
the electron gains energy which is equal to q*V
that is 0.5*mv1^2 - 0.5*mv2^2 = q*V
m mass of electron = 9.1*10^-31 kg, q = -1.6*10^-19 C
0.5mv1^2 = 0.5*mv2^2 - q*V
Given
diameterof the disls d = 1.9 cm, radius r = 0.95 cm = 0.0095m
sseparation of the disks s = 2.2mm = 2.2*10^-3m
charge q = 16 nC
from part A E = 5.64*10^8 N/C
when a proton shot from centre of the +ve disk to the centre of the -ve disk
the proton gains energy from q*V = change in k.e , proton should barely reach -ve disk
Potential V = E*d = 5.64*10^8*(2*0.0095+2.2*10^-3) V = 11956800 V
now q*v = 11956800*1.6*10^-19 J= 1.913088*10^-12 J
which is equal to the kientic energy of the proton = 0.5*mv^2
==> v = sqrt(2*1.913088*10^-12/(1.67*10^-27)) m/s
= 47865680.33 m/s
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