If the total positive charge is Q = 1.62×10^-6 C, what is the magnitude of the electric field...
Homework Answers
electric field E =k (Q/d^2) .......... (1)
where k =8.99*10^9 N.m^2/C^2
charge Q =1.62*10^-6 C
distance d =1.53 m
substitute these values in eq (1), we get
electric field E =(8.99*10^9)[(1.62*10^-6)/(1.53)^2]
E =6.22*10^3 N/C
E =6.22 kN/C
electric field E =k (Q/d^2) .......... (1)
where k =8.99*10^9 N.m^2/C^2
charge Q =1.62*10^-6 C
distance d =1.53 m
substitute these values in eq (1), we get
electric field E =(8.99*10^9)[(1.62*10^-6)/(1.53)^2]
E =6.22*10^3 N/C
E =6.22 kN/C
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caused by this charge at point P is 6220 N/C at a distance d= 1.53m
from the charge. Now find the magnitude of the force on an electron
placed at point P. Recall that the charge on an electron has
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