An Fe^3+ ion is located 400 nm (400 times 10^-9 m, about 4000 atomic diameters)

Question

An Fe^3+ ion is located 400 nm (400 times 10^-9 m, about 4000 atomic diameters) from a Cl^- ion, as shown in the figure below (ions not shown to scale). Determine the magnitude and direction of the electric field E_A^rightarrow at location A, r = 101 nm to the left of the Cl^- ion. magnitude N/C direction to the right Determine the magnitude and direction of the electric field E_B^rightarrow at location B, r = 101 nm to the right of the Cl^- ion. magnitude 1.72e4 N/C direction to the left If an electron is placed at location A, what are the magnitude and direction of the force on the electron? magnitude 6.612e4 N direction to the left

Explanation / Answer

r = 101 nm
Distance of A from Fe+3, = 400 nm - r = 400 - 101 = 299 nm

(a)
Ea = (8.9 * 10^9 * 3 * 1.6 * 10^-19) /(299 * 10^-9)^2 + (8.9 * 10^9 * 1.6 * 10^-19) /(101 * 10^-9)^2
Ea = 1.87 * 10^5 N/C
Direction :- To the Right

(b)
Distance of B from Fe+3, = 400 nm + r = 400 + 101 = 501 nm
Eb = (8.9 * 10^9 * 3 * 1.6 * 10^-19) /(501 * 10^-9)^2 - (8.9 * 10^9 * 1.6 * 10^-19) /(101 * 10^-9)^2
Eb = 1.23 * 10^5 N/C
Direction :- To the Left

(c)
Force on Electron, Fa = q*Ea
Fa = 1.6 * 10^-19 * 1.87 * 10^5 N
Fa = 2.99 * 10^-14 N
Direction :- To the Left

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