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Posted by on Wednesday, January 12, 2011 at 12:30am.

A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.18 m. The particle experiences a constant electric force, and its motion is along the line of action of the force. The difference between the particle's electric potential energy at A and B is EPEA − EPEB = +9.50 10-4 J.
(a) Find the magnitude and direction of the electric force that acts on the particle.
magnitude________ N
direction _________ (amongst, against, perpendicular to?)

(b) Find the magnitude and direction of the electric field that the particle experiences.
magnitude________N/C
direction __________ (amongst, against, perpendicular to?)

  • Physics - , Wednesday, January 12, 2011 at 8:09am

    (a) The change in potential energy divided by the distance is the force.
    F = 9.50*10^-4/0.18 Newtons
    The force direction is from B to A, because the potential energy is greater at A, and the charge is positive.

    (b) Divide the force by the charge to get the electric field strength. It will be in the same direction as the force.

  • Physics - , Sunday, September 2, 2012 at 2:53am

    wdqq

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