An electron is near a positive ion of charge +9e and a negative ion of charge ?8

Question

An electron is near a positive ion of charge +9e and a negative ion of charge ?8e (see the figure above). (Take a = 3.98 ?m, b = 3.84 ?m, and ? = 65.0?.)

A) Find the magnitude and direction of the resultant force on the electron? (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.)

B)Find the magnitude and direction of the electron's instantaneous acceleration? (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.)

An electron is near a positive ion of charge +9e and a negative ion of charge ?8e (see the figure above). (Take a = 3.98 ?m, b = 3.84 ?m, and ? = 65.0?.) A) Find the magnitude and direction of the resultant force on the electron? (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.) B)Find the magnitude and direction of the electron's instantaneous acceleration? (Let right be the +x-direction. Measure the angle counter-clockwise from the +x-axis.)

Explanation / Answer

(a) Assume the origin of the x-y axis corresponds with the position of the electron. The force of the +9e ion on the electron is: F+9e = K(9e)(- e) / (4.53E-6 m)^2 = (8.99E9 Nm^2/C^2)(- 9)(1.60E-19 C)^2 / (4.53E-6 m)^2 F+9e = -1.01E-16 N {this force points toward the +9e ion, which is the -x direction} The force of the -8e ion on the electron is: F-8e = K(-8e)(- e) / (3.00E-6 m)^2 = (8.99E9 Nm^2/C^2)(8)(1.60E-19 C)^2 / (3.00E-6 m)^2 F-8e = 2.05E-16 N {this force points away from the -8e ion, which is ? = 55.6° clockwise from the +x axis} The magnitude of the combined force, Fc, from the two ions using the law of cosines is: Fc^2 = a^2 + b^2 - 2ab cos 55.6 = (1.01E-16 N)^2 + (2.05E-16 N)^2 - 2(1.01E-16 N)(2.05E-16 N) cos55.6 = 2.88E-16 N^2 Fc = 1.70E-16 N The direction of the combined force using the law of sines is: sin 55.6 / 1.70E-16 = sin(angle) / 1.01E-16 N angle = arcsin(0.49) = 29.4° The direction of Fc is 29.4 + 55.6 = 85° clockwise from the +x axis {this puts it in the fourth quadrant} (b) Using Newton's Second Law: M A = Fc where M = mass of the electron, and A = acceleration of the electron A = Fc / M = (1.70E-16 N) / (9.11E-31 kg) = 1.87E14 m/s^2 {this is the magnitude} The direction of A is the same as Fc, 85° clockwise from the +x axis {this puts it in the fourth quadra

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