How would I calculate the magnitude of the electric field?

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Homework Help Overview

The discussion revolves around calculating the magnitude of the electric field experienced by an electron released from rest in a uniform electric field, which accelerates vertically upward over a specified distance in a short time frame.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation, Assumption checking

Approaches and Questions Raised

  • Participants explore the relationship between electric field strength and force, with one suggesting the use of the equation E = F/q. There is also a discussion about the direction of the electric field based on the electron's upward motion.

Discussion Status

Some participants have provided guidance on the nature of the electric field and the forces acting on the electron, emphasizing the need to consider both gravitational and electric forces. There is an ongoing exploration of the assumptions necessary for the calculations, particularly regarding the uniformity of the electric field.

Contextual Notes

Participants note the importance of understanding the forces at play, including the gravitational force acting on the electron, and the requirement for the electric field to be homogenous for the calculations to be straightforward.

hotmail590
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An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 m in the first 3 us (microsecond??) after it is released.

Part A What is the magnitude of the electric field?

How would I calculate the magnitude of the electric field? I am not sure which formula to use, but I am guessing E = F/q

E= F/q also equals 1/(4pi(8.854x10^-12)) * Q/r^2

If that is the right equation to use, then I know we have r already (4.50m) but how do we solve for Q?


Part B What is the direction of the electric field? Upward or Downward?


Since it says the electron moves upward after released, will that mean the direction of the electric field is upward also?


Thank you for your help.
 
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You're being asked to find the strength of the (uniform) applied electric field - not the electric field produced by the electron.
 
hotmail590, may I remind you that this kind of questions is to be posted in the homework/textbook help section of this forum ?
 
@hotmail590: You see, the electron is moving vertically upwards, it is overcoming gravity and this can only be due to another force (which is stronger than the gravitational force). Now, the key question is: what is the source of this force? You've got the answer in your problem statement: There's an electric field (which exerts a force on any charged object within its reach).
You already wrote down the right equation for the connection between the force on a charged object in an e-field and the e-field strength.
How to proceed: Clearly you must find the magnitude of the force. This comes down to a kinematics/dynamics problem: You are given information about the acceleration of the electron (it starts at rest, I suppose, and will have traveled 4.5m in 3 microseconds). Using this you can calculate the total acceleration of the electron. Note though that this acceleration is the result of two forces (from the gravitational and the electric field), so (using the mass of the electron to calculate the gravitational force) you can calculate the force due to the e-field.
Now, that's it (as you surely know the charge of an electron).
I suggest you do the math and explain all steps involved and post it all here, so that we can check.
Regards...Cliowa
Btw, one more thing: For this to work it is absolutely mandatory that the e-field be homogenous. Do you know why?
 
cliowa,

Do you know why?

Please hold off until he solves the original problem, if you get my drift! ;)
 
Tide said:
cliowa,
Please hold off until he solves the original problem, if you get my drift! ;)
Alright, I just felt this was the basic assumption which makes the calculation that straight-forward (You're implying it if your proceeding as I suggested).
 

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