Potential difference of falling charge particle

In summary, a particle with a mass of 7*10e-5 kg and a charge of 1e-6 C moves 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. The potential difference Vb-Va is unknown, but it can be found by equating the electric field and the gravitational field, using the sum of all forces and the conservation of energy equations. The work involved in this process may provide some clues.
  • #1
phymateng
6
0

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 Vb-Va?


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.
 
Physics news on Phys.org
  • #2
W = q*ΔV

So you have Δ½mv² and Δmgh.

So if there was some work involved ... any ideas?
 
  • #3


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.
 

Related to Potential difference of falling charge particle

1. What is potential difference?

Potential difference, also known as voltage, is the difference in electrical potential energy between two points in an electric field. It is measured in volts (V) and represents the amount of work that must be done to move a unit of positive charge from one point to another.

2. How is potential difference related to falling charge particles?

Potential difference is related to falling charge particles through the concept of electric potential energy. As a charged particle falls from a higher potential to a lower potential, it experiences a decrease in potential energy, which is converted into kinetic energy. This change in potential energy is what creates the potential difference.

3. How is potential difference calculated for falling charge particles?

The potential difference for falling charge particles can be calculated using the formula V = ΔPE/q, where V is the potential difference in volts, ΔPE is the change in electric potential energy in joules, and q is the charge of the particle in coulombs. Alternatively, it can also be calculated by dividing the work done on the charge particle by the amount of charge moved.

4. What factors affect the potential difference of falling charge particles?

The potential difference of falling charge particles is affected by the distance between the two points, the amount of charge on the particle, and the strength of the electric field. All of these factors can influence the amount of work done on the particle and therefore, the potential difference.

5. How is potential difference used in practical applications?

Potential difference is used in a wide range of practical applications, such as generating electricity in power plants, powering electronic devices, and creating electric circuits. It is also important in understanding and analyzing the behavior of moving charged particles in electric fields, which is crucial in fields such as particle physics and engineering.

Similar threads

  • Introductory Physics Homework Help
Replies
6
Views
409
Replies
22
Views
1K
  • Introductory Physics Homework Help
Replies
15
Views
411
  • Introductory Physics Homework Help
Replies
1
Views
206
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
321
  • Introductory Physics Homework Help
Replies
4
Views
846
  • Introductory Physics Homework Help
Replies
2
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
293
  • Introductory Physics Homework Help
Replies
23
Views
426
Back
Top