Electric Field Force Exerted Vertically

In summary: But it is important to be careful about the exponents.E = F/q = ma/qEq / m = aa = [(4*10^5) * (-1.6*10^-19)] / (1.673*10^-27)a = -3.83 m/s^2Right idea, but redo the arithmetic. (Watch those exponents.)In summary, the conversation discusses the calculation of the distance a proton is deflected downward when it enters a uniform electric field with a magnitude of 4.0 × 105 N/C and a speed of 1.5 × 107 m/s. Using the equations for electric force and acceleration, the acceleration of the proton is found
  • #1

Homework Statement


A proton enters the uniform electric field produced by the two charged plates shown below. The magnitude of the electric field is 4.0 × 105 N/C, and the speed of the proton when it enters is 1.5 × 107 m/s. What distance d has the proton been deflected downward when it leaves the plates?
1xNg1qU.png


Homework Equations


E = F/q
F = ma
PE = mgh
KE = 1/2mv^2

The Attempt at a Solution


mgh = 1/2mv^2
gh = 1/2v^2
h = v^2 / 2g

Original height: h = (1.5*10^7)^2 / 2(9.8) = 1.15*10^13 m

E = F/q = ma/q
Eq / m = a
a = [(4*10^5) * (-1.6*10^-19)] / (1.673*10^-27)
a = -3.83 m/s^2

I'm not really sure where to go from here. Would I need time to get velocity from the acceleration I found? I'm just trying to find velocity for the proton with the force of the electric field acting on it and then using that in the height equation (which I'm not sure is the right one to use here) to find the difference between the original height where the proton goes straight forward (no electric field).

I also know the force of an electric field and can use its velocity to determine the time that the proton takes to travel through the 12 cm (or 12 * 10^-6 m) space but I have no idea how to compute the vertical distance it takes depending on the downward force exerted by the field.
 

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  • #2
Civil_Disobedient said:
I also know the force of an electric field and can use its velocity to determine the time that the proton takes to travel through the 12 cm (or 12 * 10^-6 m) space but I have no idea how to compute the vertical distance it takes depending on the downward force exerted by the field.
Use the given horizontal velocity to calculate the travel time through the field. Given the time and the acceleration that you calculated, you can apply basic kinematics to find the vertical distance.
 
  • #3
Doc Al said:
Use the given horizontal velocity to calculate the travel time through the field. Given the time and the acceleration that you calculated, you can apply basic kinematics to find the vertical distance.

d = vt
t = d/v = (12*10^-2) / (1.5*10^7) = 8*10^-9s

Using kinematic y = yo + vo + 1/2at^2, I removed the yo and vo to get y = vo + 1/2at^2.
I first solved for the proton without the electric field
y = 1/2at^2
y = 1/2(9.8)(8*10^-9)^2 = 3.136*10^-16

Next, I solved for the proton with the field
y = 1/2at^2
y = 1/2(-3.83)(8*10^-9)^2 = -1.23*10^-16

So would the height difference be between 3.136*10^-16 and -1.23*10^-16? I'm not sure because these numbers look weird to me.

 
  • #4
Civil_Disobedient said:
Using kinematic y = yo + vo + 1/2at^2
That is not dimensionally consistent. You cannot add a distance to a velocity. What have you left out?
 
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  • #5
haruspex said:
That is not dimensionally consistent. You cannot add a distance to a velocity. What have you left out?

That's just a basic kinematic equation, I may have incorrectly remembered it. I ended up using y=1/2at^2 to calculate height.
 
  • #6
Civil_Disobedient said:
E = F/q = ma/q
Eq / m = a
a = [(4*10^5) * (-1.6*10^-19)] / (1.673*10^-27)
a = -3.83 m/s^2
Right idea, but redo the arithmetic. (Watch those exponents.)
 
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  • #7
Doc Al said:
Right idea, but redo the arithmetic. (Watch those exponents.)

Ah, I forgot to write in exponents for acceleration. a = 3.83*10^13. I also missed out how charge of a proton is positive, so acceleration would be positive.

So from there I recalculated y=1/2at^2
y = 1/2 * (3.83*10^13) * (8*10^-9) = 0.00123m

I'm a bit confused on finding the height difference, don't I also need to find height of the proton moving straight forward without the field? But when I do that, I still have 3.136*10^-16m.
 
  • #8
Civil_Disobedient said:
I'm a bit confused on finding the height difference, don't I also need to find height of the proton moving straight forward without the field?
The force that the field exerts on the proton is so much greater than its weight that you can safely ignore such a tiny correction due to gravity.
 
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  • #9
Doc Al said:
The force that the field exerts on the proton is so much greater than its weight that you can safely ignore such a tiny correction due to gravity.

I see, that makes sense now. Thank you so much!
 
  • #10
Civil_Disobedient said:
That's just a basic kinematic equation, I may have incorrectly remembered it.
Quite so. The v0 term should be v0t, but since you used v0=0 you got away with it.
 
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1. What is an electric field force exerted vertically?

An electric field force exerted vertically is a type of force that exists between two charged particles or objects, where one is located above the other. This force is caused by the interaction of the electric fields of the two particles or objects.

2. How is the strength of an electric field force exerted vertically calculated?

The strength of an electric field force exerted vertically is calculated using Coulomb's Law, which states that the force between two charged particles or objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

3. What factors affect the magnitude of an electric field force exerted vertically?

The magnitude of an electric field force exerted vertically is affected by the magnitude of the charges of the two particles or objects, the distance between them, and the medium in which they are located. The force increases with increasing charge and decreases with increasing distance between the particles or objects. The type of medium also affects the force, as some materials can weaken or enhance the electric field.

4. How does the direction of an electric field force exerted vertically vary?

The direction of an electric field force exerted vertically depends on the relative charges of the particles or objects. If the charges are opposite, the force will be attractive and will pull the particles or objects towards each other. If the charges are the same, the force will be repulsive and will push the particles or objects away from each other.

5. What are some real-life applications of electric field force exerted vertically?

Electric field force exerted vertically has many practical applications, including in capacitors, where it is used to store electrical energy, and in electrostatic precipitators, which remove particles from air or gas streams. It is also important in the formation of lightning, and is utilized in technologies such as Van de Graaff generators and cathode ray tubes.

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