A particle with a charge of 8.01 Times 10^-10 C and a mass of 6.0 Times 10^-28 k

Question

A particle with a charge of 8.01 Times 10^-10 C and a mass of 6.0 Times 10^-28 kg moves uniformly with a speed of 24000.0 in a circular orbit around a stationary particle of charge 4.806 Times 10^-19 C. Calculate the radius of the orbit. R = m A charge q_1 = 0 .4 C is 3 m from a charge q_2 = 0 .8 C. What is the magnitude of the electrostatic force on q_1? Be sure to include units in your answer. Answer: 5 mu C charge is placed at the origin. An identical charge is placed 6.2 in from the origin on the x axis. A third identical char, placed 6 .2 m from the origin on the y axis What is the magnitude of the net electrostatic force on the charge at the origin? (Don't enter a sign for your force: just the magnitude) Be sure to include units in your answer. Answer

Explanation / Answer

Charge q = -8.01 x10 -19 C

Mass m = 6x10 -28 kg

Speed v = 24000 m/s

Stationary charge Q = 4.806 x10 -19 C

We know coulomb's force between q and Q = centripeatl force

    KqQ / r 2 = mv 2 / r

   KqQ / r = mv 2

From this r = KqQ/(mv 2)

                  = (8.99 x10 9 )(8.01 x10 -19 )(4.806 x10 -19 ) /[(6 x10 -28 )(24000 2) ]

                  = 1.001 x10 -8 m

(2). q1 = 0.4 C

     q2 = 0.8 C

      r = 3 m

Required force F = Kq1q2/r 2

                         =(8.99x10 9)(0.4)(0.8) / 3 2

                         = 319.64 x10 6 N/C

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