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A particle has a charge of +1.5 µC and moves from point A to point B, a distance of 0.25 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 = +6.50x10^-4 J.

(a) Find the magnitude and direction of the electric force that acts on the particle.

(b) Find the magnitude and direction of the electric field that the particle experiences.

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Category:- Physics
Reference No.:- M91016590

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