Electricity, Conducting Plates and Magnetic Fields

AI Thread Summary
A pair of square parallel conducting plates, spaced 0.01 meters apart and connected to a 200-volt power supply, creates an electric field of 2000 N/C directed downward. An electron entering this region with a horizontal speed of 3 x 10^7 m/s experiences an upward acceleration of 3.513 x 10^14 m/s² due to the electric field. The vertical displacement of the electron while between the plates is calculated to be approximately 0.0505 meters. To maintain a horizontal trajectory, a magnetic field must be applied, requiring calculations to balance the electric and magnetic forces acting on the electron. The discussion highlights the need for careful consideration of forces and parameters to achieve zero net force on the electron.
Zeus
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Howdie all,
I'm soo glad I found this forum, people are genuinely working together instead of ignoring their fellow physics brethren =)

A pair of square parallel conducting plates, having sides of length .05 meter, are .01 meter apart and are connected to a 200 volt power supply. An electron is moving horizontally with a speed of 3 x 10^7 m/s when it enters the region between the plates. Neglect gravitation and the distortion of the electric field around the edges of the plates.
-----------------------------------------------------------------
+
---
200 V -----> -e 3 x 10^7 m/s
---
-
-----------------------------------------------------------------
Side length is .05 m

1. Determine the magnitude of the electric field in the region between the plates and indicate its direction on the figure above.

E=-V/d
E=200/.1
E=2000, down
-----------------------------------------------------------------
+ !
---!
200 V -----> -e 3 x 10^7 m/s!
---!
- vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv (down arrows)
-----------------------------------------------------------------


2. Determine the magnitude and direction of the acceleration of the electron in the region between the plates.

E=F/q F=ma

Eq=F
2000 x 1.60 x 10^-19=ma
a=3.513 x 10^14 m/s^2, up


3. Determine the magnitude of the vertical displacement of the elctron for the time interval during which it moves through the region between the plates.

d=vt
.05=3 x 10^7 x t
t=1.67 x 10^-9
d=vt +(1/2)a x t^2
d=.0505 m


4.On the diagram below, sketch the path of the electron as it moves through and after it emerges from the region between the plates. The dash lines in the diagramhave been added for reference only.
-----------------------------------------------------------/////
.005/////////////////////////////////////////////////////////
/////////////////////////////
(median line)----------------------------------------------

.005
------------------------------------------------------------
(The line is curving slowly, then faster upward)
5. A magnetic field could be placed in the region between the plates which would cause the electron to continue to travel horizontally in a straight line through the region between the plates. Determine both the magnitude and the direction of this magnetic field.

This is really the one I don't know how to do. I realize that the vertical force must be zero between the magnetic and the electrical force, but I don't know how to do it mathematically.

Thanks, again, for your help =)
 
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1) List all of the forces acting on the electron (there are two). This part is physics.

2) Determine which ones you are allowed "to play with." (There is only one.) This part is engineering.

3) Find all possible arrangements of them that give you what you know you need: zero net force. Choose one (there is still only one). This part is algebra.

4) Realize how these forces are related to certain parameters in the problem (i.e. magnetic field, velocity, plate separation, electric field, ... not all of the ones I mention are relevant). This part is physics.

5) Determine which ones you are allowed "to play with." This part is engineering.

6) Find all possible arrangements of them that give you what you know you need: determined from step 3. Choose one. This part is algebra.
 
Reply to Turin

Ok, I've given your post some thought:
F=Magnitude of Force of Magnetic field
E=Electric field
theta=@

F=qvBsin(@)

E=-v/d

0=F + E
0=1.60 x 10^-19 x 2000 x B x sin90 - 200/.1
0=(3.2 x 10^-16)B - 2000
B=6.25 x 10^18

Was this the idea?
Thanks Turin
 


Originally posted by Zeus
Was this the idea?
Roughly. I would recommend using a capital V for voltage and a lower case v (or even something else, like u) for velocity. I think it might have gotten you into trouble and you may want to double check. Also, you cannot add apples and oranges, or, in this case, E-fields and forces. You need to calculate the electric force before you add it to the magnetic force. I will go ahead and work the calculation to see if I get the same answer and then let you know.

My answer does not resemble yours.
 
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