Work done by a uniform Ring of Charge (due tomorrow plz help

1. Feb 12, 2016

Melo out

1. The problem statement, all variables and given/known data
A ring of diameter 7.90 cm is fixed in place and carries a charge of 5.20 μC uniformly spread over its circumference.
How much work does it take to move a tiny 3.40 μC charged ball of mass 1.50 g from very far away to the center of the ring?

2. Relevant equations
V=(KQ)/r^2
intergral of k(Qq)/r^2 *ds= work done
work done= - (change in potential energy)
work done= F dot delta S (otherwise known as displacement)

3. The attempt at a solution
First i realized that force=Eq and that the force felt would be changing due to how far away the charge was from the uniform circle. So i setup
intergral of K(Qq)/r^2*ds from r to infinity
i got -KQq/r as my indefinite intergral, which i thought was the electric potential.energy
(K(3.4*10^-6)*(5.2*10^-6))/0.079= 2.01= electric potential energy ( at infinity it would be 0).

2. Feb 13, 2016

cnh1995

3. Feb 13, 2016

haruspex

It's rather hard to tell what you did without more details. What expression did you use to relate r to s? Did you take into account which component of the field you needed?

4. Feb 13, 2016

2nafish117

you have a typo in relevant equations
V=kQ/r not(kQ/r^2)

yes the force on a charge q is Eq,but the E which you have taken is only due to a point charge not a charged ring!!!!!!!!!
there are two ways to approach this problem
1.find the electric field on the axis of the ring as a function of how far the test charge is away from the centre of the ring.
then integrate to find the work you have put in to move the charge from infinity to the centre of the ring.

2.the second method is a bit more straightforward. find the potential as a function of how far the test charge is from the centre of the ring .
then finding the work done is easy.