Recent content by Alex Z

Awesome, thank you very much.

F = (k*q*q)/(r^2) = ((8.9*10^9)*(15*10^-6)*(2*10^-6)) / (0.0487^2) = 112.58 N So from there, I can use F = m*a to find the acceleration: a = F / m = (112.58) / (4*10^-3) = 28144.5 m/s^2 I believe that should be the final answer.

Initial Kinetic Energy = 1/2*(4*10^-3)*(40^2) = 3.2 Final Kinetic Energy = 1/2*(4*10^-3)*(23^2) = 1.058 Oops. So that means that rf = 0.0487 meters. I guess the values weren't great enough to make a significant difference.

Well I believe that the equation here would be Uei + Ki = Uef + Kf So (k*q*q)/0.08 + 3.44 = (k*q*q)/rf + 1.137 With all this equated, rf should be equal to 0.0487 meters. From there, my first instinct would be to find the change in distance and then use kinematics.

I calculated the force using everything was given to me. Since the radius of the sphere wasn't given, I just assumed that the size of the sphere was negligible and I therefore didn't have to take it into account. As for the location of the sphere, I assume conservation of energy needs to be used...

Homework Statement A point charge q1=15.00μC is held fixed in space. From a horizontal distance of 8.00 cm , a small sphere with mass 4.00×10−3kg and charge q2=+2.00μC is fired toward the fixed charge with an initial speed of 40.0 m/s . Gravity can be neglected. What is the acceleration of the...