If the total positive charge is Q = 1.62×10−6 C, what is the magnitude of the...
If the total positive charge is Q
= 1.62×10−6 C, what is the magnitude of the electric field caused
by this charge at point P, a distance d = 1.53 m from the
charge?
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
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