How Do You Calculate the Electric Field in a Capacitor from Electron Deflection?

[Xels]

Homework Statement

The figure shows an electron entering a parallel-plate capacitor with a speed of 5.4×10⁶m/s. The electric field of the capacitor has deflected the electron downward by a distance of 0.618cm at the point where the electron exits the capacitor.

Find the magnitude of the electric field in the capacitor.

Homework Equations

f=ma
f=k(q/r²)

The Attempt at a Solution

I'm trying to figure out how to gauge the force applied from the field, which I'm assuming can be taken from the displacement & speed traveled. I'm also assuming that gravity isn't a factor here, so if E=f/q₀ where am I getting my f from?

Relevant data:
length of capacitor: 2.25E-2m
e^-_m: 9.11E-31kg
e^-_q: 1.60E-19C
e^-_v: 5.4E6m/s
$$\theta$$:15.36$$\circ$$

 

[cepheid]

You're getting your f from basic dynamics (Newton's laws of motion). In a parallel plate capacitor, the electric field is uniform (constant in magnitude and direction). So, throughout its journey across the capacitor plates, the electron has a constant vertical force applied to it. Therefore, it experiences a constant vertical acceleration (vertical meaning perpendicular to its direction of motion). You know how long it takes it to traverse the plates, because you know its horizontal speed (which won't change) and you know the length of the plates. Call this time interval "t". So, the question reduces to: a particle undergoes a displacement d in a time t when moving with constant acceleration a. What must "a" have been in order for it to have traveled that far in that amount of time? Once you know a, you know f.

This is no different from a projectile motion problem. Only the nature of the force (electrostatic vs. gravitational) has changed.

 

[Xels]

Great, that's exactly the kick I needed. Thank you.

 

[darwizzl]

so what is exactly the answer?

 

[asteriatic]

I would like to clarify a few things before proceeding with my response. First, can you provide the diagram or figure mentioned in the problem? It would be helpful in visualizing the situation. Also, can you clarify what the angle of 15.36 degrees represents in this problem?

Assuming that the angle represents the deflection of the electron from its initial path, we can use the equation F=ma to calculate the force exerted on the electron by the electric field. We can rewrite this equation as F=qE, where q is the charge of the electron and E is the electric field strength.

To find the electric field strength, we need to know the charge of the electron and the force exerted on it. We can calculate the force by using the formula F=ma, where m is the mass of the electron and a is the acceleration caused by the electric field. We can determine the acceleration by using the formula a=v^2/r, where v is the initial velocity of the electron and r is the radius of curvature of its path.

Using the given data, we can calculate the radius of curvature of the electron's path to be 0.618cm or 0.00618m. Plugging in the values for mass, velocity, and radius of curvature, we get a=5.4E6^2/0.00618=5.87E12 m/s^2.

Now, we can calculate the force exerted on the electron by using the formula F=qE. Substituting the values for charge and acceleration, we get F=(1.60E-19)*(5.87E12)=9.39E-7 N.

Finally, we can calculate the electric field strength by using the formula E=F/q. Substituting the values for force and charge, we get E=9.39E-7/1.60E-19=5.87E12 N/C.

Therefore, the magnitude of the electric field in the capacitor is 5.87E12 N/C. It is important to note that this calculation assumes a uniform electric field between the parallel plates of the capacitor. If the electric field is not uniform, then this calculation may not be accurate. Additionally, the presence of other forces, such as gravity, may also affect the electron's path and thus the calculated electric field strength.