# What is the velocity of the particle when the particle reaches 3R?

The upper half of the ring with radius R is +Q, the lower half is -Q.

On it's axis, There are Q particles at a distance of 10R from the center.

What is the velocity of the particle when the particle reaches 3R?

Q=5C

R=6cm

q=2C

m=2.0*10^-6 kg

k=9*10^9 N*m^2/C^2

V0=5.0*10^4 m/s

### 1 Answer

- Steve4PhysicsLv 71 month agoFavourite answer
EDIT

If we assume the object can’t move in the y-direction then the velocity remains constant (5.0*10^4 m/s).

That's because of symmetry. The x-components of attraction and repulsion forces (from the upper and lower halves of the ring) have equal magnitudes and opposite directions, so they cancel-out. No x-force means no x-acceleration.

(The forces' y-components have the same direction, so do not cancel. And to complicate matters, once the particle moves in the y-direction, the x-components are no longer equal magnitudes.)

END EDIT

That doesn’t make sense. There is a downwards (-y direction) force. So the particle will move in a curved path, and not pass though the point marked 3R.

Steve, can you look at that too ?

https://answers.yahoo.com/question/index?qid=20200530230709AA2nwsC