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25 November, 01:57

A proton at speed v = 3.00 * 105 m/s orbits at radius r = 1.00 cm outside a charged sphere. Find the sphere's charge.

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  1. 25 November, 02:16
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    q = 1.04*10^-9C

    Explanation:

    The force on a charge orbiting at the center of the sphere (Fc) is must be equal to the electrostatic force acting on the charge. (Fq)

    Fc = Fq

    The force that causes a body to bev attracted towards the center of the sphere is the centripetal force.

    Centripetal force (Fc) = ma = mv²/r where m is the mass of the body

    v is the speed

    r is the velocity

    The electrostatic force acting on the body is derived according to the Coulomb's law of electrostatic attraction.

    Fq = kqe/r²

    Where k is the constant

    e is the electron orbiting outside the sphere

    r is the distance between the charge and the electron

    Since fc=fq,

    mv²/r = kqe/r²

    Making the charge the subject of the formula we have;

    mv²r = kqe

    q = mv²r/ke

    Given the mass of the charge =

    8.84*10^-12kg

    v = 3.0*10^5

    r = 1.0cm = 0.01m

    k = 1/4Π£o = 1/4Π*1.67*10^-27

    k = 4.77*10^-27

    e = 1.6*10^-19C

    Substituting the values in the formula given,

    q = 8.84*10^-12 * (3.0*10^5) ² (0.01) / (4.77*10^-27) * 1.6*10^-19

    q = 1.04*10^-9C

    The sphere's charge is 1.04*10^-9
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