charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric field at a poin

Question

charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric field at a point on the y-axis 0.500m from the origin.

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Thiên Thanh 2 months 2021-07-30T09:54:16+00:00 1 Answers 2 views 0

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    2021-07-30T09:55:19+00:00

    Answer:

    Explanation:

    Electric field due to charge at origin

    = k Q / r²

    k is a constant , Q is charge and r is distance

    = 9 x 10⁹ x 5 x 10⁻⁶ / .5²

    = 180 x 10³ N /C

    In vector form

    E₁ = 180 x 10³ j

    Electric field due to q₂ charge

    = 9 x 10⁹ x 3 x 10⁻⁶ /.5² + .8²

    = 30.33 x 10³ N / C

    It will have negative slope θ with x axis

    Tan θ = .5 / √.5² + .8²

    = .5 / .94

    θ = 28°

    E₂ = 30.33 x 10³ cos 28 i – 30.33 x 10³ sin28j

    = 26.78 x 10³ i – 14.24 x 10³ j

    Total electric field

    E = E₁  + E₂

    = 180 x 10³ j +26.78 x 10³ i – 14.24 x 10³ j

    = 26.78 x 10³ i + 165.76 X 10³ j

    magnitude

    = √(26.78² + 165.76² ) x 10³ N /C

    = 167.8 x 10³  N / C .

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