A point charge with a charge q1 = 3.80 μC is held stationary at the origin. A second point charge with a charge q2 = -4.10 μC moves from the point x= 0.140 m , y=0 to the point x= 0.240 m , y= 0.240 m. How much work is done by the electric force on q2?

Question

I know that work = force*distance, and that force = k*q1*q2 / r^2, but I don't know what r is and I don't know how to calculate the distance given both change in the x and y directions. What should I do? Thanks

Answer 1

Hard way: do an integral force dot distance over the path, and of course the force magnitude and direction (cosTheta) is dependent on position. That would be an interesting exercise.

Easy Way. Calculate the difference in potential energies at each point: that is the work it took to get there, INDEPENDENT of path.
Work= Vb*q2 - Va*q2
and for course, V=kq1/distance
distance= distance from the point to the origin.

Answer 2

For the latter method, I understand the concept of path independence and computing work using the two points, such that (U = potential energy)
W = Uf - Ui
And I understand that V (electric potential) is
V = U/q0
such that U = Vq
And that V is also equal to kq/r

But I need a bit of help understanding the difference between Vf and Vi, and why both are multiplied by q2. Also, what distances are used? Initially, it should be .14m in the x direction; but the final distance using the pythagorean theorem is .34m at some angle. Is the angle irrelevant?

Answer 3

vf=final point
vi=voltage at initial point
final distance does not depend on angle. Voltage potential is a Scalar, not a vector.