A +2.3 mC charge is placed 35 cm away from a stationary -3.4 mC charge. What is

Question

A +2.3 mC charge is placed 35 cm away from a stationary -3.4 mC charge. What is the electrical potential? In a 15 Volt potential, a particle has a potential energy of -0.45 J. What is the particle's charge? A -1.5 mC charge is moving towards a stationary +6.8 mC charge. What is the change in potential energy as the particle moves from 75 cm away from the stationary charge to only 33 cm from the stationary charge? Did the potential energy increase or decrease? A +1.5 mC charged particle (m = 4.3 kg) is placed 1.2 m from a +4.4 mC stationary charge. If it starts from rest, how fast will the particle be traveling when it is 2.0 m away from the stationary charge? A +2.7 mC charged particle (m = 1.5 kg) is shot with an initial velocity of 202 m/sec towards a +1.8 mC stationary charge. If the particle starts out 1.4 meters from the stationary charge, how close will it come to the charge before turning around and moving away? A proton (m = 1.7 times 10^-27 kg, q = +1.6 times 10^-19 C) is placed at edge of the positive plate of a 2.4 times 10^-6 F capacitor. If the capacitor holds 17 mC of charge on its positive plate, how fast will the proton be moving when it reaches the negative plate of the capacitor?

Explanation / Answer

Given:

Charge on 1st particle = q1=2.7 x 10-3 C

mass of 1st particle = m1=1.5 kg

initial velocity of 1st particle = v1= 202 m/s

final velocity of 1st particle at the closest position=v2=0

initial distance of 1st particle from stationary charge= r1= 1.4 m

final distance of 1st particle from stationary charge at the closest position = r2=?

charge of stationary charge =q2 = 1.8 x 10 -3 C

In this case, the change in kinetic energy of the charged particle= increase in potential energy of the system

Change in kinetic energy of charged particle = 1/2 x m1 x (v12 - v22 )

= 1/2 x 1.5 x (2022 - 0)

= 30603 J

Initial potential energy of the system =kq1q2/r1

=9 x 109 x 2.7 x 10-3 x 1.8 x 10 -3 / 1.4

= 31242 J

final potential energy of the system =kq1q2/r2

= 9 x 109 x 2.7 x 10-3 x 1.8 x 10 -3 / r2

= 43740/r2

Change in potential energy of system = 43740/r2 - 31242

= 30603

so, r2= 0.7 m

so the partcile will come close to a distance of 0.7m before turning around and moving away.

all the best in the course work

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