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27 Nov 2019
A particle has a charge of +1.5 µC and moves from point A topoint B, a distance of 0.25 m. The particle experiences a constantelectric force, and its motion is along the line of action of theforce. The difference between the particle's electric potentialenergy at A and B is EPEA - EPEB = +6.50x10^-4 J.
(a) Find the magnitude and direction of the electric forcethat acts on the particle.
magnitude in N?
direction ? along against perpendicular to
(b) Find the magnitude and direction of the electric field that theparticle experiences.
magnitude in N/C ?
direction ? along against perpendicular to
A particle has a charge of +1.5 µC and moves from point A topoint B, a distance of 0.25 m. The particle experiences a constantelectric force, and its motion is along the line of action of theforce. The difference between the particle's electric potentialenergy at A and B is EPEA - EPEB = +6.50x10^-4 J.
(a) Find the magnitude and direction of the electric forcethat acts on the particle.
magnitude in N?
direction ? along against perpendicular to
(b) Find the magnitude and direction of the electric field that theparticle experiences.
magnitude in N/C ?
direction ? along against perpendicular to
Sixta KovacekLv2
25 Oct 2019
