Question

D. Imagine a negatively charged particle (9--0.2 C) free to move back and forth along a straight line. There are other nearby charged particles that create electric potential and electric field at all points on the line. The system is made up of the negatively charged particle and the other particles that created the potential and field. As usual, we will assume an isolated system, so that this system's total energy is constant. The particle starts at x = 0 and is given a very quick push. This force does work on the particle, so that it has 4 J of kinetic energy at x - and begins to move to the right. This is the only particle in the system that moves. Here are some data on the electric potential that the positively charged particle foels at various points along the line: V-0 Volts x1 cm V-20 Volts X-2 cm V-40 Volts x=3 cm V-35 Volts X-4 cm V-30 Volts x = 5 cm V - 25 Volts Important to remember. U. = 9V, where is the potential created by the other particles and q is the charge of the particle that moves back and forth along the line. 1. Atx=0 cm, the system's potential energy is J, the system's total energy J. and the particle's kinetic energy is 2. Atx - 2 cm, the system's potential energy is J, the system's total energy is J, and the particle's kinetic energy is 3. As the particle moved from 0 to x = 2 cm, (moving in the positive direction) it was (speeding up, slowing down, maintaining a constant speed] 4. Therefore, the electric force it felt was positive, negative, zero). 5. Therefore, the electric field between x 0 and x 4 cm was (positive, negative, zero). Remember, this is a negative charge. 6. This is exactly what we expect, because the direction of decreasing potential is the (positive, negative direction, and electric field always points in the direction of increasing, decreasingl potential. 7. At x = 5 cm, the system's potential energy is J, the system's total energy is ____J, and the particle's kinetic energy is 8. As the particle moved from x 2 cm to x = 5 cm, (moving in the positive direction) it was (speeding up, slowing down, maintaining a constant speed]. 9. Therefore, the electric force it felt was positive, negative, zero). 10. Therefore, the electric field between x 2 cm and x = 5 cm was positive, negative, zero). Remember, this is a negative charge.

this is one question with many parts, thanks for the help!

Answer 1

-2 (93-4 1 = -0.20) the -+ -3 XO 1 2 3 4 Ov 20v to 35V 30v 25 v you at x=0 qu -0.2x 40 = -85 Enngy 3 Potentine Energy 0xco= qV = 9x0 = 0 KE = 1 0 2 = 45 Total Energy = U+KE = 0 + 45 = UJ at x=2 {v=uou] Potential Energy Us=2 Total E will remain same = UT E = P. E + KE 4 = 8+K E KIE=12] As particle move x=2 tox 2m (K. E increasing It means speeding up 9 dierection of electric field towards lef 18 & change is 10 to Xz2! Potential is increasing 24 means regatione. Direction of force towards exe (the) ② (Potential Increasing -> Dineetion of pettenkok Eleemic freled well always higher to Lowerpotential] Direction of Electric field will be towards - we sanis. 6 Decreasing potential - Kephes Direction of electr'e freed always hynin potential to lower potential . 2 x=5 Totala Energy particle a Ux-s = qu = -0.2x 25 = -55 remains same = 45 K. E = EU = 4+5=95 moved x22 to 5 Kinetic Energy is decreasing c showing down) Direction of Electric field will be towards the x-axis due decrease in potential. then force will be towards - re x-anis. forom point (q) Electric field towards thue xasis © (10