Anonymous
Anonymous asked in Science & MathematicsPhysics · 4 months ago

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

Attachment image

1 Answer

Relevance
  • 4 months ago
    Favourite 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.

Still have questions? Get answers by asking now.