Relationship of particle charge to speed?

[temptasian]

Homework Statement

The four points shown in the attached picture are near a positively charged rod (shaded circle). Points W and Y are equidistant from the rod, as are points X and Z. A charged particle with mass 3.0 x 10^-8 kg is released from rest at point W and later observed to pass point X.
Suppose the magnitude of the charge is 2 microcoloumbs and that the speed of the particle is 40 m/s as it passes point X. Suppose that a second particle with the same mass as the first but NINE times the CHARGE were released from rest at point W. Would the speed of the second particle as it passes point X be greater than, less than, or equal to the speed of the first particle (with charge of 2 microcoulombs) as it passes point X?

Homework Equations

From KE = 1/2 (mv^2), I know that charge is not involved here, and since mass is the same, I believe the speed of the second particle is equal to the speed of the first particle.
However, I'm not sure if I should be using this equation to make a relationship between speed and charge. Other possible equations to use may be: U(potential energy) = qV, and W = -qEd = ΔU
Am I on the right track?

The Attempt at a Solution

 

[BruceW]

yes, you need to use the 'other possible equations'. You've got ΔU=qΔV This is the change in potential energy, so what will be the change in kinetic energy? Also, is ΔV the same or different in the two situations?

 

[temptasian]

ΔV is constant for both situations. So if I use ΔU = qΔV, when charge increases, ΔU has to increase also, which means kinetic energy has decreased. Since KE has decreased, and we know that KE = 1/2 mv^2, the speed of the second particle will be less than the speed of the second particle. I hope I'm correct here?

 

[BruceW]

Yes, the ΔV is the same in both situations.

So if I use ΔU = qΔV, when charge increases, ΔU has to increase also, which means kinetic energy has decreased. Since KE has decreased, and we know that KE = 1/2 mv^2, the speed of the second particle will be less than the speed of the second particle.

But this isn't right. Think about the sign of ΔU. The particle is being pushed away from the rod, so how does the potential change when the particle moves away from the rod?

 

[asteriatic]

As a scientist, it is important to always consider all relevant equations and variables in a problem. In this case, the relationship between particle charge and speed can be understood through the principles of electrostatics and classical mechanics.

First, let's consider the equation for electric potential energy, U = qV, where q is the charge of the particle and V is the electric potential. From this equation, we can see that the potential energy of a charged particle is directly proportional to its charge. This means that a particle with a larger charge will have a higher potential energy than a particle with a smaller charge.

Next, let's consider the equation for work done by an electric field, W = -qEd, where E is the electric field strength and d is the distance the particle moves in the field. From this equation, we can see that the work done by the electric field on a particle is directly proportional to its charge. This means that a particle with a larger charge will experience a greater force from the electric field and therefore, will have a greater acceleration and speed.

So, in the given scenario, the second particle with nine times the charge of the first particle will have a higher potential energy and experience a greater force from the electric field. This will result in a higher acceleration and ultimately, a higher speed as it passes point X.

In conclusion, there is a direct relationship between particle charge and speed. A particle with a larger charge will have a greater potential energy and experience a greater force from the electric field, resulting in a higher speed.