For a positively charged particle at the origin and anegatively charged particle

Question

For a positively charged particle at the origin and anegatively charged particle lying to its right on the positive xaxis, the electrostatic force on the negative is attractive, andCoulomb's law gives the magnitude of that force:F=k|q1||q2|/x2. Theconstant k has a value of 8.99 x 109N*m2/c2, q1 and q2represent the electric charges (with units of Coulombs, C) and x isthe distance between the charges. Assume of positivly chargedparticle with q1=0.025 C is fixed at the origin andattracts a negatively charged particle with q2=-0.050 C.(a) How much work does the positive charge do on the negativecharge as their seperation decreases from 0.035 m to 0.014 m? (b)Assume the negative charge starts at rest and has a mass of 0.40kg. What is the speed at the end of the journey? (a)_____ J (b)_____ m/s Please show steps and formulas used....thank you For a positively charged particle at the origin and anegatively charged particle lying to its right on the positive xaxis, the electrostatic force on the negative is attractive, andCoulomb's law gives the magnitude of that force:F=k|q1||q2|/x2. Theconstant k has a value of 8.99 x 109N*m2/c2, q1 and q2represent the electric charges (with units of Coulombs, C) and x isthe distance between the charges. Assume of positivly chargedparticle with q1=0.025 C is fixed at the origin andattracts a negatively charged particle with q2=-0.050 C.(a) How much work does the positive charge do on the negativecharge as their seperation decreases from 0.035 m to 0.014 m? (b)Assume the negative charge starts at rest and has a mass of 0.40kg. What is the speed at the end of the journey? (a)_____ J (b)_____ m/s Please show steps and formulas used....thank you

Explanation / Answer

        a.   Workdone   =   Final p.e. of twocharges   -   initial p.e. of twocharges             W   =   -k * q1 * q2 /x2   -   ( - k *q1 * q2 / x1)                   =   k * q1 * q2 * (1 /x1   -   1/x2)                   =   8.99* 109 * 0.025 * ( - 0.050) * (1 /0.035   -   1/0.014)                   =   4.82* 108   J    b.   According to work energytheorem,   k.e.    =   W       (1/2) * m *v2   =   W    0.5 * 0.40 *v2   =   4.82 *108    v   =   (4.82* 108 / 0.20)          =   4.91 *104   m/s    v   =   (4.82* 108 / 0.20)          =   4.91 *104   m/s

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