a) Find the magnitude of the electric field at the position of this charge.
b)What is the direction of the electric field at the position of this charge?
F = Eq
3.2 = E*(-7*10^-6)
E = -3.2/(7*10^-6)
E = -0.457*10^6 N/C
Direction :- away from charge
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Answers & Comments
F = Eq
3.2 = E*(-7*10^-6)
E = -3.2/(7*10^-6)
E = -0.457*10^6 N/C
Direction :- away from charge