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17 September, 20:00

A dipole is centered at the origin, and is composed of charged particles with charge and, separated by a distance 4 * 10-10 m along the axis. The charge is on the axis, and the charge is on the axis. A proton is located at m. What is the force on the proton, due to the dipole? N An electron is located at m. What is the force on the electron, due to the dipole? N

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  1. 17 September, 22:38
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    a) F_net = 2.883*10^-15 N ~ 0 N on proton @ m

    b) F_net = 1.442*10^-15 N ~ 0 N on an electron @ position m

    Explanation:

    Given:

    - + e position = m

    - - e position = m

    - Charge of electron e = - 1.602 * 10^-19 C

    - Charge of a proton p = 1.602 * 10^-19 C

    - Coulomb's constant k = 8.99 * 10^9

    Find:

    - F_net on a proton @ position m

    - F_net on an electron @ position m

    Solution:

    - Find the force F_e due to + e on proton using Coulomb's Law:

    F_e = k * (+e) * (+e) / r^2

    where, r = (4*10^-8 + 2*10^-10) = 4.02*10^-8 m

    F_e = 8.99 * 10^9 * (1.602 * 10^-19) ^2 / (4.02*10^-8) ^2

    F_e = 1.4277*10^-13 N

    - Find the force F_-e due to - e on proton using Coulomb's Law:

    F_-e = k * (+e) * (-e) / r^2

    where, r = (4*10^-8 - 2*10^-10) = 3.98*10^-8 m

    F_-e = - 8.99 * 10^9 * (1.602 * 10^-19) ^2 / (3.98*10^-8) ^2

    F_-e = - 1.45652711*10^-13 N

    - F_net is the sum of F_e and F_-e as follows:

    F_net = F_e + F_-e

    F_net = 10^-13 * (1.4277 - 1.45652711)

    F_net = 2.883*10^-15 N = 0 N

    - Find the force F_e due to + e on electron using Coulomb's Law:

    F_e = k * (+e) * (-e) / r^2

    where, r^2 = ((4*10^-8) ^2 + (2*10^-10) ^2) = 1.60004*10^-15 m^2

    F_e = - 8.99 * 10^9 * (1.602 * 10^-19) ^2 / 1.60004*10^-15

    F_e = 1.4419622*10^-13 N

    - Find the force F_-e due to - e on electron using Coulomb's Law:

    F_-e = k * (-e) * (-e) / r^2

    where, r^2 = ((4*10^-8) ^2 + (2*10^-10) ^2) = 1.60004*10^-15 m^2

    F_e = - 8.99 * 10^9 * (1.602 * 10^-19) ^2 / 1.60004*10^-15

    F_e = - 1.4419622*10^-13 N

    - F_net is the sum of F_e and F_-e as follows:

    F_net = F_e*sin (Q) + F_-e*sin (Q) = 2*F_e*cos (Q)

    where, Q is the angle between x-axis and r

    sin (Q) = 2*10^-10 / sqrt (1.60004*10^-15)

    sin (Q) = 5.0*10^-3

    Hence, F_net = 2 * (1.4419622*10^-13) * (5.0*10^-3)

    F_net = - 1.442*10^-15 N = 0 N
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