a)Find the magnitude of the electric force F exerted on a point charge + 6e by a

Question

a)Find the magnitude of the electric force F exerted on a point charge + 6e by a point charge – 3 e which is located 7.90 nm away.
Express the result in unit [N], and to three significant figures. If you must use scientific notation, please enter as follows: ex. 0.00012 = 1.2E-4. Only answer in numerical values, no units.



b)An electron is released into a uniform electric field of magnitude 923 N/C. By what factor exceeds the acceleration of the electron its gravitational acceleration?

Express the result to three significant figures; it has no units. If you must use scientific notation, please enter as follows: e.g. 0.000123 = 1.23E-4. Only answer in numerical values, no units. For negative numbers, leave no space between the negative sign and the number: e.g. right: -1.00, wrong: - 1.00.

Explanation / Answer

a) Given Magnitude of two charges are q1 = +6e = 6(1.6 *10-19 C) = 9.6 E-19 C And q2 = -3e = -3(1.6 *10-19 C) = 4.8 E-19 C Distance of seperation, r =7.90 nm Magnitude of force is F = k q1q2 /r2    = (9.0 E+9 Nm2/C2) (9.6 E-19 C) (4.8 E-19 C) / (7.90 E-9 m)2 F = 6.645 E-11 N ------------------------------------------------------------------------------- b) Uniform electric field of magnitude , E = 923 N/C      Charge of the electron ,q = 1.6 E-19 C      Mass of the electron ,m = 9.11 E-31 kg      From Newton's second law F = ma     But F = qE.so,         ma = qE           a = (1.6 E-19 C) (923 N/C) / (9.11E-31 kg)           a = 1.62 E+14 m/s2 Acceleration due to gravity , g = 9.8 m/s2    Now, a /g = (  1.62 E+14 m/s2) / 9.8 m/s2 a /g = 1.65 E +13 So, acceleration of the electron is 1.65 E +13 times the gravitational acceleration Acceleration due to gravity , g = 9.8 m/s2    Now, a /g = (  1.62 E+14 m/s2) / 9.8 m/s2 a /g = 1.65 E +13 So, acceleration of the electron is 1.65 E +13 times the gravitational acceleration

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