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28 June, 01:54

Two charges are located in the xx-yy plane. If q1=-4.10 nCq1=-4.10 nC and is located at (x=0.00 m, y=1.080 m) (x=0.00 m, y=1.080 m), and the second charge has magnitude of q2=3.60 nCq2=3.60 nC and is located at (x=1.20 m, y=0.600 m) (x=1.20 m, y=0.600 m), calculate the xx and yy components, ExEx and EyEy, of the electric field, E⃗ E→, in component form at the origin, (0,0) (0,0). The Coulomb force constant is 1 / (4π?0) = 8.99*109 N⋅m2/C21 / (4πϵ0) = 8.99*109 N⋅m2/C2.

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  1. 28 June, 04:07
    0
    Due to first charge, electric field at origin will be oriented towards - ve of y axis.

    magnitude

    Ey = - 8.99 x 10⁹ x 4.1 x 10⁻⁹ / 1.08² j

    = - 31.6 j N/C

    Due to second charge electric field at origin

    = 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.2²+.6²

    = 8.99 x 10⁹ x 3.6 x 10⁻⁹ / 1.8

    = 18 N/C

    It is making angle θ where

    Tanθ =.6 / 1.2

    = 26.55°

    this field in vector form

    = - 18 cos 26.55 i - 18 sin26.55 j

    = - 16.10 i - 8.04 j

    Total field

    = - 16.10 i - 8.04 j + ( - 31.6 j)

    = - 16.1 i - 39.64 j.

    Ex = - 16.1 i

    Ey = - 39.64 j.
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