A particle of charge Q is fixed at the origin of an xy coordinate system. At t = 0 a particle (m = 0.693 g, q = 3.09 µC is located on the x axis at x = 24.6 cm, moving with a speed of 74.0 m/s in the positive y direction. For what value of Q will the moving particle execute circular motion? (Neglect the gravitational force on the particle.)

The moving particle will execute a circular motion when the Electric force between the particles is equal to the centripetal force necessary for the circular motion at a velocity equal to 74m/s and a radius = 0.246m. You can find the value for the centripetal force with this expression:

[tex]F_c = m\frac{v^2}{r} = 6.93*10^{-4}kg\frac{(74m/s)^2}{0.246m} = 15.43N[/tex]

Now, this is the value that the electric force between the charges should be. The electric force is given by this expresion:

[tex]F_e = k*\frac{Qq}{r^2}[/tex]

k is the Coulomb constant, equal to 9*10^9 Nm^2/C^2. Then:

[tex]F_e = k*\frac{Qq}{r^2}\\Q = \frac{F_e r^2}{k*q} \\Q = \frac{15.43 N*(0.246m)^2}{9*10^9 Nm^2/C^2*3.09*10^{-6}C} = 33.57 *10^{-6}C[/tex]

Or 33.57μC