A point charge q1=+2.40μC is held stationary at the origin. A second point charge q2=−4.30μC moves from the point x=0.125m , y=0, to the poi

Question

A point charge q1=+2.40μC is held stationary at the origin. A second point charge q2=−4.30μC moves from the point x=0.125m , y=0, to the point x=0.275m , y=0.275m.

Part A

What is the change in potential energy of the pair of charges?

Express your answer in joules to three significant figures.

Part B

How much work is done by the electric force on q2?

Express your answer in joules to three significant figures.

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Thiên Di 5 months 2021-08-26T11:58:49+00:00 1 Answers 2 views 0

Answers ( )

    0
    2021-08-26T12:00:29+00:00

    Answer:

    Explanation:

    Potential energy of a pair of charges

    = k q₁q₂ /r₁₂

    At r₁₂ = .125 m

    = – 9 x 10⁹ x 2.4 x 4.3 x 10⁻¹² / .125

    = – 743 x 10⁻³J

    At r₁₂ = √2 x .275 m

    – 9 x 10⁹ x 2.4 x 4.3 x 10⁻¹² / √2 x .275

    = – 238.85  x 10⁻³J

    = 239 x 10⁻³ J

    Change = – 238.85  x 10⁻³ +743 x 10⁻³

    = 504.15 x 10⁻³ J

    B )

    Work done by electric force will be same but in negative sign

    = – 504.15 x 10⁻³ J

    = – 504 x 10⁻³ J

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