A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is the magnitude

Question

A +15 nC point charge is placed on the x axis at x = 1.5 m, and a -20 nC charge is placed on the y axis at y = -2.0m. What is the magnitude of the electric field at the origin?

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Thanh Thu 6 months 2021-07-31T19:37:33+00:00 1 Answers 27 views 0

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    2021-07-31T19:39:19+00:00

    Answer:E=75\ N/m

    Explanation:

    Given

    First charge of q_1=15\ nC is placed at x=1.5\ m

    Second charge  q_2=-20\ nC is placed at y=-2\ m

    Electric field is given by

    E=\frac{kq}{r^2}

    Electric field due to q_1 is away from it

    E_1=\frac{9\times 10^9\times 15\times 10^{-9}}{(1.5)^2}

    E_1=60\ N/m

    Electric field due to q_2

    E_2=\frac{9\times 10^9\times 20\times 10^{-9}}{2^2}

    E_2=45\ N/m

    Net electric field will be vector addition of two

    \vec{E_{net}}=\vec{E_1}+\vec{E_2}

    \vec{E_{net}}=-60\hat{i}-45\hat{j}

    Magnitude of Electric field is

    E=\sqrt{60^2+45^2}

    E=75\ N/m

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Giải phương trình 1 ẩn: x + 2 - 2(x + 1) = -x . Hỏi x = ? ( )