Imagine two free electrons that collide elastically in an acidic solution where one electron was moving and the other electron w
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Answer:
E = 1.19 N/C
Explanation:
Let's first determine the length of the arc which can be given as:
L= Rθ
where:
L = length of the arc
R = radius of curvature
θ = angle in radius
L = (9.09×10⁻²m)(2.59)
L = (0.0909)(2.59)
L = 0.235431 m
Then, the magnitude of electric field that Q produces at the center of curvature can be calculated by using the formula:
Since
where;
L = length
Q = charge
λ = density of the charge;
then substituting for λ, we have :
substituting our given parameter; we have:
E = 1.1889 N/C
E = 1.19 N/C
∴ the magnitude of the electric field that Q produces at the center of curvature = 1.19 N/C