Potential difference of falling charge particle

[phymateng]

Homework Statement

A particle of mass 7*10e-5 kg and charge 1e-6 C from point A to point B a distance of 3 m in the Earth's gravitational field. The kinetic energy of the particle decreases by 5.3mJ during this movement. The acceleration of gravity is 9.8 m/s^2. What is the potential difference V_b-V_a?

Homework Equations

I don't know how to get the Electric Field. I know that its a positive charge going in the same direction as the gravitational field. Since the problem says that the final kinetic energy is decreasing, then I know the velocity is decreasing and giving me a hint that there is an electric field counteracting the gravitational field. I could use the sum of all forces to equate both the fields, but that is only when the charge would stop. Do I have the right Idea?

The Attempt at a Solution

Tried to do the sum of all forces and the conservation of energy.

 

[LowlyPion]

W = q*ΔV

So you have Δ½mv² and Δmgh.

So if there was some work involved ... any ideas?

 

[nlightner]

I would approach this problem by first recognizing that the particle is experiencing two types of forces: gravitational force and electric force. The gravitational force, given by Fg = mg, is acting downwards on the particle while the electric force, given by Fe = qE, is acting upwards in the opposite direction of the gravitational force.

To determine the potential difference between point A and B, we need to consider the change in potential energy of the particle. From the conservation of energy, we know that the change in potential energy is equal to the negative of the change in kinetic energy, given by ΔU = -ΔK. Therefore, we can write:

ΔU = -ΔK
mgΔh = -ΔK
(7*10^-5 kg)(9.8 m/s^2)(3 m) = -(-5.3*10^-3 J)
2.058*10^-3 J = 5.3*10^-3 J
ΔU = 5.3*10^-3 J

Next, we can use the formula for potential energy in a uniform electric field, given by U = qEd, where d is the distance between the two points A and B. We can rearrange this equation to solve for the electric field, E:

E = U/qd

Substituting the values given in the problem, we get:

E = (5.3*10^-3 J)/(1*10^-6 C * 3 m) = 1.767*10^3 N/C

Since we know that the electric field is acting in the opposite direction of the gravitational field, we can use the electric field value to determine the potential difference between point A and B:

Vb - Va = -Ed = -(1.767*10^3 N/C)(3 m) = -5.301 V

Therefore, the potential difference between point A and B is -5.301 V, meaning that the potential at point B is lower than the potential at point A. This makes sense since the particle is losing kinetic energy as it moves from point A to B, indicating that the potential at point B is lower due to the presence of an electric field counteracting the gravitational field.