The direction of the electric field at point P is +x direction

[imationrouter03]

Thank you for viewing this,.. here's my question i hope u can help:

A ring-shaped conductor with a radius 2.30 has a total positive charge 0.125 uniformly distributed around it. The center of the ring is at the origin of coordinates O.
(a)The magnitude of the electric field at point P, which is on the positive x-axis at 39.0 is 7.35 N/C
(b)The direction of the electric field at point P is +x direction

here's the question:
A particle with a charge of 2.60 is placed at the point P described in part (a). What is the magnitude of the force exerted by the particle on the ring?

and also What is the direction of the force exerted by the particle on the ring?

Thank you for your time,.. any advice would be appreciated

 

[AKG]

The force on a point charge is equal to the product of the charge and the electric field at the point, i.e.:

\vec{F} = q\vec{E} = (2.60\ C)(7.35\ N/C\ \hat{i}) = 19.11\ N\ \hat{i} = 19.1\ N\ \hat{i}

Not that \hat{i} is the unit vector in the x direction, and the last bit, where I want from 19.11 to 19.1 was to take care of significant digits.

 

[eJavier]

And by Newton's 3rd Law, the force exerted by the particle on the ring is the same in magnitude but of opposite sign of what AKG computed.