A uniform electric field with a magnitude of 250 N/C is directed in the positive x direction. A 12 μC charge moves from the origin to the po

Question

A uniform electric field with a magnitude of 250 N/C is directed in the positive x direction. A 12 μC charge moves from the origin to the point (20.0 cm, 50.0 cm). What is the change in the electrical potential ener- gy of the system as a result of the change in position of this charge?

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Thái Dương 3 days 2021-07-19T16:41:09+00:00 1 Answers 2 views 0

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    2021-07-19T16:42:22+00:00

    Answer:

    -0.6 mJ

    Explanation:

    We first find the change in electric potential ΔV from

    ΔV = -∫E.dr = -Ercosθ were E = magnitude of electric field = 250 N/C and r = distance of charge from the origin = √[(20 – 0)² + (50 – 0)²] = √2900 = 53.58 cm = 0.536 m. The direction is θ = tan⁻¹(50/20) = 68.2 which is the angle between E and r.

    So ΔV = -Ercosθ = – 250 N/C × 0.536 m × cos68.2 = -49.77 V

    The electrical potential energy ΔU = qΔV were q = care = 12 μC = 12 × 10⁻⁶ C

    ΔU = qΔV = 12 × 10⁻⁶ C × -49.77 V = – 597.2 × 10⁻⁶ J = – 0.5972 mJ ≅ -0.6 mJ

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