Two spherical point charges each carrying a charge of 40 μC are attached to the two ends of a spring of length 20 cm. If its spring constant

Question

Two spherical point charges each carrying a charge of 40 μC are attached to the two ends of a spring of length 20 cm. If its spring constant is 120 Nm-1 , what is the length of the spring when the charges are in equilibrium?

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Thiên Hương 7 months 2021-07-12T11:26:40+00:00 1 Answers 0 views 0

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    2021-07-12T11:28:04+00:00

    Answer:

    x=3

    Explanation:

    From the question we are told that:

    Charge Q=40 \mu C

    Length L=20cm=0.20m

    Spring constant k=120Nm^{-1}

    Generally the equation for Force between Charges is mathematically given by

     F=k\frac{q_1 q_2}{r^2}

     F=9*10^9\frac{40*10^{-6}^2}{0.2^2^2}

     F=360N

    Therefore

     F=kx

     x=\frac{F}{k}

     x=\frac{360}{120}

     x=3

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