Consider the potential field V(x, y) which is 0 and -Vo(Vo>0

In summary: No.In summary, the equation for the particle's velocity in a region of zero potential energy is ##v=v_0+E##, where v_0 is the velocity in the original region of zero potential energy.
  • #1
pallab
36
3

Homework Statement


Consider the potential field V(x, y) which is 0 and -Vo (V0 > 0) respectively in the regions
of y greater and less than zero . Let θ and θ' be the angles of incidence and refraction of
the particle with the y-axis at the point of incidence as it crosses the x-axis . The ratio
sin(θ) / sin(θ ') is given (in terms of Δ = Vo / E) by
region1
upload_2017-4-5_19-53-33.png
region2
(a)√(1+2Vο/E)
(b)√(1+Vο/E)
(c)1+Vο/E
(d)1+2Vο/E

Homework Equations


nsinθ=n'sinθ'
sinθ/sinθ'=v/v'[where v=velocity of particle in 1 & v'in 2]

The Attempt at a Solution


1/2mv2=E=1/2mv'2-Vο
1/2mv'2-1/2mv2=Vο
(v'2 -v2)/v2=2Vο/mv2
(v'2 /v2 )-1=Vο/(1/2mv2)
∴sinθ'/sinθ=√(1+Vο/E)

Which is similar to option b but in question it is sinθ/sinθ'
 
Last edited:
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  • #2
pallab said:
sinθ/sinθ'=v/v' [where v=velocity of particle in 1 & v'in 2]
Are you sure this equation is correct? If Vo is positive, which would be larger: v or v'? θ or θ'?
 
  • #3
I have just used snell's law.
θ∝v
but I am not sure about increase or decrease of kinetic energy in the presence of potential +ve or -ve.
 
  • #4
TSny said:
Are you sure this equation is correct? If Vo is positive, which would be larger: v or v'? θ or θ'?
I have just used snell's law.
θ∝v
but I am not sure about increase or decrease of kinetic energy in the presence of potential +ve or -ve.
 
  • #5
This problem deals with a particle moving from a region of zero potential energy to another region of constant potential energy -Vo.

For this situation, Snell's law as written for light ##\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}## does not apply. Have you covered how Snell's law is modified for the particle situation?
 
Last edited:
  • #6
TSny said:
This problem deals with a particle moving from a region of zero potential energy to another region of constant potential energy -Vo.

For this situation, Snell's law as written for light ##\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}## does not apply. Have you covered how Snell's law is modified for the particle situation?
No.
TSny said:
This problem deals with a particle moving from a region of zero potential energy to another region of constant potential energy -Vo.

For this situation, Snell's law as written for light ##\frac{\sin \theta_1}{v_1} = \frac{\sin \theta_2}{v_2}## does not apply. Have you covered how Snell's law is modified for the particle situation?
No.
 
  • #7
Then you will need to derive "Snell's law" for the particle.

As a start, consider the x and y components of the velocity of the particle.
Does vx change when the particle passes through the origin?
Does vy change?
 

1. What is the meaning of the potential field V(x, y)?

The potential field V(x, y) represents the potential energy of a particle at a specific point (x, y) in a two-dimensional space. It is a scalar function that measures the work required to move a particle from one point to another in the presence of a force field.

2. Why is the potential field V(x, y) defined as 0 and -Vo?

In this case, the potential field V(x, y) is defined as 0 at one point and -Vo at another point because it represents a region of space where the potential energy is constant. This is typically seen in cases where the force field is conservative, meaning that the work done by the force only depends on the initial and final positions, not the path taken.

3. What does Vo represent in the potential field V(x, y)?

Vo represents the potential energy at a specific point in the field. It is a constant value and is typically chosen to be a positive number for convenience. The value of Vo can affect the overall shape and behavior of the potential field.

4. How is the potential field V(x, y) related to the force field in a two-dimensional space?

The potential field V(x, y) and the force field in a two-dimensional space are related by the gradient operator. The force field is equal to the negative gradient of the potential field, meaning that the force at a specific point is in the direction of steepest decrease of the potential energy.

5. Can the potential field V(x, y) be used to calculate the trajectory of a particle in the two-dimensional space?

Yes, the potential field V(x, y) can be used to calculate the trajectory of a particle in the two-dimensional space by using the principle of conservation of energy. Since the potential energy is constant along a trajectory, the total energy of the particle (kinetic + potential) remains constant, and the trajectory can be calculated using equations of motion.

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