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27 Nov 2019

A particle of charge +3.0 x 10-8 is in a unifrom electric fielddirected to the left. It is released from rest and moves a distanceof 5.0 cm and its kinetic energy is increased by +4.5 x -4 J.

a) What is the work fone by the electric force? units?

b) What is the magnitude of the electric field? units?

c)What is the electric potential of the starting point relative tothe endpoint of the path? units?

Use symbols for the units.

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Sixta Kovacek
Sixta KovacekLv2
9 Jul 2019

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Related textbook solutions

Physics: Principles with Applications
7th Edition, 2013
Giancoli
ISBN: 9780321625922

Related questions

A particle of charge +3.0 x 10-8 is in a unifrom electric fielddirected to the left. It is released from rest and moves a distanceof 5.0 cm and its kinetic energy is increased by +4.5 x -4 J.

a) What is the work fone by the electric force?______________units?_____________

b) What is the magnitude of the electric field?_______________units?____________

c)What is the electric potential of the starting point relative tothe endpoint of the path____________? units?________________

Use symbols for the units.

Review. Two parallel plates having charges of equal magnitude but opposite sign are separated by 12.0 cm. Each plate has a surface charge density of 36.0 nC/m2. A proton is released from rest at the positive plate. Determine

(a) the magnitude of the electric field between the plates from the charge density

(b) the potential difference between the plates

(c) the kinetic energy of the proton when it reaches the negative plate

(d) the speed of the proton just before it strikes the negative plate

(e) the acceleration of the proton

(f) the force on the proton

(g) From the force, find the magnitude of the electric field.

(h) How does your value of the electric field compare with that found in part (a)?

1. A particle has a charge of +1.5μC and moves from point A topoint B, a distance of 0.25m. 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 EPE(a) - EPE(b) = +6.00 x 10^-4 J.

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

magnitude __________N direction____________

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magnitude___________N/C direction___________

2. A positive charge +q1 is located to the left of a negativecharge -q2. On a line passing through the two charges, there aretwo places where the total potential is zero. The first place isbetween the charges and is 1.0 cm to the left of the negativecharge. the second place is 5.2 cm to the right of the negativecharge.

(a) What is the distance between the charges?

___________cm

(b) Find q1/q2, the ratio of the magnitudes of the charges.

q1/q2 = ____________

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