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21 January, 03:56

It is known that the electric force of repulsion between two electrons is much stronger than the gravitational attraction. For two electrons at a distance d apart, calculate the ratio of the size of the electrostatic repulsion to that of the gravitational attraction. Use the following dа ta: k = 8.99x10^9 Nm2/C2, e = 1.60x10^-19 C, G = 6.67x10^-11 Nm2/kg2, me = 9.11x10^-31 kg.

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  1. 21 January, 04:16
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    Answer: 348.22*10^ (-104)

    Explanation: gravitational force between electrons is gotten using formulae below

    Fg = G * me*me/d²

    Fg = gravitational force

    G = gravitational constant = 6.67x10^-11 Nm2/kg2

    me = mass of electron = 9.11*10^-31 kg

    d = distance between electrons

    Fg = (6.67x10^-11 * 9.11*10^-31 * 9.11*10^-31) / d²

    Fg = 64.04*10^ (-72) / d²

    Fe = k*e * e/d²

    Where Fe = strength of electric field

    K = electric constant = 8.99x10^9 Nm²/C²

    e = 1.60x10^-19 C

    Fe = 8.99x10^9 * 1.60x10^-19 * 1.60x10^-19/d²

    Fe = 2.23*10^ (-28) / d²

    We are to find Fe / Fg

    =2.23*10^ (-28) / d² : 64.04*10^ (-72) / d²

    =2.23*10^ (-28) / d² * d²/64.04*10^ (-72)

    = 2.23*10^ (-28) / 64.04*10^ (-72)

    = 348.22*10^ (-104)
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