Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force acting on 1

Question

Three charges 1.5*10-6, 3*10-6, -3*10-6 are placed at three vertices of an equilateral triangle of side 30cm. Find the net force acting on 1.5*10-6 due to the other two

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Thu Nguyệt 7 months 2021-07-30T15:42:22+00:00 2 Answers 2 views 0

Answers ( )

    0
    2021-07-30T15:43:33+00:00

    Answer:

    Explanation:

    Check attachment for free body diagram.

    Fnet =F12 + F13

    Therefore,

    The force of attraction between two charges is given as

    F12 = kq1q2/r²

    F12=9×10^9×1.5 ×10^-6×3×10^-6/0.3²

    F12 = 0.45N

    Also,

    F13 = kq1q3/r²

    F13=9×10^9×1.5 ×10^-6×3×10^-6/0.3²

    F13 = 0.45N

    From the diagram we notice that F12 is acting between positive x-axis and y axis

    F12 = F12Sin30 •i + F12 Cos30 •j

    F12 = 0.45Sin30 •i + 0.45Cos30 •j

    F12 = 0.225 •i + 0.39 •j N

    F13 is acting between positive x axis and negative y axis

    Then,

    F13 = F12Sin30 •i — F12 Cos30 •j

    F13 = 0.45Sin30 •i — 0.45Cos30 •j

    F13 = 0.225 •i — 0.39 •j N

    Then,

    Fnet = F12 + F13

    Fnet = 0.225 •i + 0.39 •j + N0.225 •i – 0.39 •j N

    Fnet = 0.45 •i N

    The net force has a magnitude of 0.45N and it is acting on the positive x axis

    Then it direction is

    X = arctan(y/x)

    X = arctan(0)

    X = 0°

    0
    2021-07-30T15:44:18+00:00

    Answer:

    F = 0N

    Explanation:

    The force between two charges is given by

    F=k\frac{q_1q_2}{r^2}

    where r is the distance between the charges and K is the Coulomb’s constant

    (k=8-89*10^9Nm^2/C^2)

    The force in the first charge is only the sum of the forces due to the other charges. Hence we have

    F_T=F_1+F_2=k\frac{q_2q_1}{r^2}+k\frac{q_3q_1}{r^2}

    F_T=(8.89*10^9\frac{Nm^2}{C^2})\frac{(3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}+(8.89*10^9\frac{Nm^2}{C^2})\frac{(-3*10^{-6}C)(1.5*10^{-6}C)}{(0.3m)^2}\\\\F_T=0.445N-0.445N=0N

    Ft=0N

    Hope this helps!!

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