1. A particle has a charge of +1.5C and moves from point A to point B, a distanc
ID: 1707266 • Letter: 1
Question
1. A particle has a charge of +1.5C and moves from point A to point B, a distance of 0.25m. 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 EPE(a) - EPE(b) = +6.00 x 10^-4 J.
(a) Find the magnitude and direction of the electric force that acts on the particle.
magnitude __________N direction____________
(b) Find the magnitude and direction of the electric field that the particle experiences.
magnitude___________N/C direction___________
2. A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two places where the total potential is zero. The first place is between the charges and is 1.0 cm to the left of the negative charge. the second place is 5.2 cm to the right of the negative charge.
(a) What is the distance between the charges?
___________cm
(b) Find q1/q2, the ratio of the magnitudes of the charges.
q1/q2 = ____________
Explanation / Answer
1. Given Charge of the particle is q = +1.5 *10^-6 C Distance moved by the particle is ,d = 0.25 m EPE(a) -EPE(b) = 6.00 *10^4J since the difference in the potential energy is equal to the work done by the electric field EPE(a) -EPE(b) =Wab so, Wab = 6.00 *10^4J (a) From the relation W =Fd F = (6.00 *10^4J) /0.25 m = 2.4 *10^5 N The direction of force is along the motion of the particle (b) And since F = qE 2.4 *10^5 N = (+1.5 *10^-6 C) E E = 1.6 *10^11 N/C The directionof electric field will be along the direction of force Post only one question for single postRelated Questions
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