Electrons in a uniform electric field.

AI Thread Summary
The discussion focuses on an experiment involving electrons in a uniform electric field created by two parallel plates, one positively charged and the other negatively charged. The experiment aims to determine the distance between the plates and calculate the strength of the electric field using a graph derived from the electron's motion. The anode voltage is set at 3,000 volts, and the relationship between the electric field strength and distance is expressed through the equation E = V/d. The graph equation provided suggests a quadratic relationship, indicating that the distance between the plates may not be necessary for calculations. The experiment's conditions, such as whether it was conducted in a vacuum or gas, are also questioned, as they could affect the results.
lord_blurg
Messages
10
Reaction score
0

Homework Statement



Basically, I am writing up a practical experiment dealing with Electrons in a uniform electric field (I.e. Using an electron gun).

There are two parallel plates, one with a positive charge and one with a negative charge. Between these to plates is a grid which allows for easy graphing.

It can be assumed that the electric field between the plates is uniform. A grid (with lines separated by 1 cm) is shown. I need to use this information to determine the distance, d, between the plates and to calculate the strength of the uniform electric field, E.

How do I find out the distance between the 2 parallel plates? (Probably simplistic, it's just that I am doing this experiment based on an experiment video I watched)

The anode voltage is set to 3,000 volts.
The graph I got was y=(.0504)(x)^1.763

Homework Equations



I am assuming this equation is related...

E = V/d (V is the potential difference between the plates.)
 
Physics news on Phys.org
Theoretically, the equation is y=(a/2/v^2)x^2

where, a is the acceleration=eE/m
v is the initial velocity, assumed to be parallel to the plates

If you substitute all these quantities to above equation then
y=eE/(2mv^2)x^2

and while mv^2/2=eV with V is the voltage

y=(E/4V)x^2

You do not need the distance between 2 plates.

Here you got the power of about 1.7<2. Hmm, was the experiment held in vacuum or in some gas?
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top