Deflection of an Electron and Capacitor problems

In summary: The 2.76 microF capacitor will have a charge of 12.3 volts and the 6.21 microF capacitor will have a charge of 5.7 volts.
  • #1
Moxin
24
0
k, Here's the problem:

An electron is accelerated horizontally from rest in a television picture tube by a potential difference of 24600 V. It then passes between two horizontal plates 6.08 cm long and 1.42 cm apart that have a potential difference of 236 V. At what angle theta (in degrees) will the electron be traveling after it passes between the plates?

-----------

Being that I have no clue how to EVEN APPROACH this problem, I just did random stuff. I have a feeling this problem involves the arctan function.. so, first I took the arctan of 1.42/6.08 - Wrong. Then the Arctan of E (which I calculated to be 16164) over the potential difference which accelerated the particle - Wrong. I tried variations of those two - Wrong. I honestly don't have the slightest clue where to go next


-----------------------------


My next problem goes:

A 2.76 microF capacitor is charged to 1230 V and a 6.21 microF capacitor is charged to 480 V. These capacitors are then disconnected from their batteries, and the positive plates are now connected to each other and the negative plates are connected to each other. What will be the potential difference across each? What will be the charge on the 2.76 microF capacitor? What will be the charge on the 6.21 microF capacitor?


-------------


Once again, I have no clue how to approach. This is absolutely depressing. I spent an hour try'na figure out each of these problems (like I do with the rest of the 15 or so, unfortunately, Booo to bad books and even worse professors). Anyways, any suggestions.. ?
 
Physics news on Phys.org
  • #2
The electron is starting from rest and being deflected in 2 perpendicular directions. Break the problem into 2 parts, 1 part per direction.

The electron starts with lots of potential energy caused by the voltage and ends with lots of kinetic energy; in both directions. Just find the final velocity in each direction then use trig to find the angle.

Ep = Ek
 
Last edited:
  • #3
Apply
[tex]\frac{mv_f^2-mv_i^2}{2}=eV[/tex]
Which is shorter way
 
  • #4


Can you clarify how I'd apply that further himanshu

Originally posted by ShawnD
The electron is starting from rest and being deflected in 2 perpendicular directions. Break the problem into 2 parts, 1 part per direction.

The electron starts with lots of potential energy caused by the voltage and ends with lots of kinetic energy; in both directions. Just find the final velocity in each direction then use trig to find the angle.

Ep = Ek

Well, before you edited and expounded step by step, I had followed what you said but still couldn't get it right...


I found in the book an example that had an equation velocity = sqrt(-2qV/m)

I applied this equation to both the horizontal direction by using 24600 V, then the vertical direction by using 236 V.. I came out with 93008330 m/s for the horizontal and 9109830 for the vertical component.. used arctan(vertical/horizontal) and came out with an angle of 5.594 degrees...and its still wrong

I don't know what else to try
 
  • #5
And any suggestions on the 2nd problem ? ?
 

1. What is the deflection of an electron and how does it relate to capacitors?

The deflection of an electron refers to the movement of an electron in an electric or magnetic field. This movement can be influenced by capacitors, which are electronic components that store electrical charge and can create electric fields that can deflect electrons.

2. How do you calculate the deflection of an electron in a capacitor?

The deflection of an electron in a capacitor can be calculated using the equation F = Eq, where F is the force on the electron, E is the electric field strength, and q is the charge of the electron. This force causes the electron to accelerate and deflect towards the positive plate of the capacitor.

3. What factors affect the deflection of an electron in a capacitor?

The deflection of an electron in a capacitor can be affected by the strength of the electric field, the distance between the plates of the capacitor, and the charge of the electron. It can also be influenced by the presence of other electric or magnetic fields in the surrounding environment.

4. How can you use deflection of electrons to measure the capacitance of a capacitor?

The deflection of electrons can be used to measure the capacitance of a capacitor by applying a known electric field and measuring the resulting deflection of the electrons. By manipulating the electric field strength and measuring the deflection, the capacitance of the capacitor can be calculated using the equation C = q/E, where C is the capacitance, q is the charge of the electron, and E is the electric field strength.

5. What are some real-world applications of deflection of electrons and capacitors?

The deflection of electrons and capacitors have various applications in electronics and technology. They are used in cathode ray tubes (CRTs) to create images on screens, in particle accelerators to manipulate and deflect charged particles, and in electronic devices such as televisions and computer monitors. Capacitors are also used in power supplies and electronic circuits to store and regulate electrical energy.

Similar threads

Replies
4
Views
332
  • Introductory Physics Homework Help
Replies
5
Views
890
  • Introductory Physics Homework Help
Replies
18
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
1K
  • Introductory Physics Homework Help
Replies
30
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
2K
  • Introductory Physics Homework Help
Replies
3
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
690
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
21
Views
2K
Back
Top