A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.00×10−4C is

Question

A positive point charge q1 = +5.00 × 10−4C is held at a fixed position. A small object with mass 4.00×10−3kg and charge q2 = −3.00×10−4C is projected directly at q1. Ignore gravity. When q2 is 4.00m away, its speed is 800m/s. What is its speed when it is 0.200m from q1?

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Kim Cúc 1 year 2021-09-03T16:21:35+00:00 1 Answers 50 views 0

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    2021-09-03T16:23:04+00:00

    Answer:

    Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

    Explanation:

    Energy conservation law: In isolated system the amount of total energy remains constant.

    The types of energy are

    1. Kinetic energy.
    2. Potential energy.

    Kinetic energy [tex]=\frac{1}{2} mv^2[/tex]

    Potential energy =[tex]\frac{Kq_1q_2}{d}[/tex]

    Here, q₁= +5.00×10⁻⁴C

    q₂=-3.00×10⁻⁴C

    d= distance = 4.00 m

    V = velocity = 800 m/s

    Total energy(E) =Kinetic energy+Potential energy

                          [tex]=\frac{1}{2} mv^2[/tex]+ [tex]\frac{Kq_1q_2}{d}[/tex]

                         [tex]=\frac{1}{2} \times 4.00\times 10^{-3}\times(800)^2 +\frac{9\times10^9\times 5\times10^{-4}\times(-3\times10^{-4})}{4}[/tex]

                        =(1280-337.5)J

                        =942.5 J

    Total energy of a system remains constant.

    Therefore,

    E [tex]=\frac{1}{2} mv^2[/tex] + [tex]\frac{Kq_1q_2}{d}[/tex]

    [tex]\Rightarrow 942.5 = \frac{1}{2} \times 4 \times10^{-3} \times V^2 +\frac{9\times10^{9}\times5\times 10^{-4}\times(-3\times 10^{-4})}{0.2}[/tex]

    [tex]\Rightarrow 942.5 = 2\times10^{-3}v^2 -6750[/tex]

    [tex]\Rightarrow 2 \times10^{-3}\times v^2= 942.5+6750[/tex]

    [tex]\Rightarrow v^2 = \frac{7692.5}{2\times 10^{-3}}[/tex]

    [tex]\Rightarrow v= 1961.19[/tex]   m/s

    Therefore the speed of q₂ is 1961.19 m/s when it is 0.200 m from from q₁.

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