an electron with initial speed of 29×10^5m/s is fired in the same direction as that of uniform electric field of magnitude of 80N/C. how far the electron travelling before stopping?
Answers
Answered by
0
Answer:
nksksucshsuufhjskidufdh
Answered by
0
Answer:
mass of electron = 9.11×10^-31kg
charge on electron q=1.6×10^-19C
initial speed Vi= 29×10^5m/s
Vf=0m/s
E= 80N/C
Explanation:
As F=qE
F=(1.6×10^-19)×80
F=1.28×10^-17C
also F=ma
compare ma=qE
a=qE/m
a=1.28×10^-17/9.11×10^-31
a=1.4×10^13
To determine distance s
2as=Vf^2-Vi^2
S=Vf^2-Vi^2/2a
S=0-(29×10^5)^2/2×1.4×10^13
S=8.41×10^12/4.28×10^13
S=1.96×10^-1
S=0.196m
Similar questions