Calculating the charge to mass ratio

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SUMMARY

This discussion focuses on calculating the charge-to-mass ratio (q/m) of a small object projected into a uniform electric field between parallel plates. The object has an initial speed of v0 = 5.00×103 m/s and is subjected to an electric field of E = 700 N/C. The object is deflected downward by a distance of d = 1.15 cm after passing through the field region. The participants outline methods for determining the vertical acceleration and the final velocity after exiting the electric field, emphasizing the importance of keeping variables symbolic until the final calculation.

PREREQUISITES
  • Understanding of electric fields and forces (E=F/q)
  • Knowledge of kinematic equations (xf=xi+vi*t+(1/2)at2)
  • Familiarity with the concept of charge-to-mass ratio (q/m)
  • Basic algebra for manipulating equations and solving for variables
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  • Learn how to derive the vertical acceleration of an object in an electric field.
  • Study the implications of keeping variables symbolic in physics problems.
  • Explore the relationship between electric force and motion in uniform fields.
  • Investigate the effects of varying initial speeds on charge-to-mass ratio calculations.
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Students in physics, particularly those studying electromagnetism, as well as educators and tutors assisting with problems related to electric fields and motion.

WestTXcollegekid
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Homework Statement


A small object with mass m, charge q, and initial speedv0 = 5.00×10^3 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1) .The electric field between the plates is directed downward and has magnitude E = 700 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.15 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.

Homework Equations


E=F/q
xf=xi+vi*t+(1/2)at^2

The Attempt at a Solution



Since gravity and air resistance air are not factors, I thought the only thing affecting the motion of the object is the electric field. Using the horizontal velocity, I calculated the time that the object was between the plates. Then, using that and the second equation I calculated the vertical acceleration of the object. From there, I expanded the first equation into E=(m*a)/q and plugged in the acceleration and given electric field and solved for q/m
 

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I don't see a question. Did you obtain a result or are you just outlining an approach you tried? How did it turn out? Did it go off the rails somewhere?
 
WestTXcollegekid said:

Homework Statement


A small object with mass m, charge q, and initial speedv0 = 5.00×10^3 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1) .The electric field between the plates is directed downward and has magnitude E = 700 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.15 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.

Homework Equations


E=F/q
xf=xi+vi*t+(1/2)at^2

The Attempt at a Solution



Since gravity and air resistance air are not factors, I thought the only thing affecting the motion of the object is the electric field. Using the horizontal velocity, I calculated the time that the object was between the plates. Then, using that and the second equation I calculated the vertical acceleration of the object. From there, I expanded the first equation into E=(m*a)/q and plugged in the acceleration and given electric field and solved for q/m
Your method, as far as you describe it, sounds right. Not clear what you did in regard to the travel after the plates. Please post your working. Please keep it symbolic (no plugging in numbers) until the final step.
 
The objective is to find the charge to mass ratio, I have attached a screenshot of the problem.
 
The issue I'm having is, how do i go about finding the y velocity after it exits the E field. and how do I find the acceleration on the object by the E field if i don't know the displacement during its time in the E field.
 
WestTXcollegekid said:
The issue I'm having is, how do i go about finding the y velocity after it exits the E field. and how do I find the acceleration on the object by the E field if i don't know the displacement during its time in the E field.
Let the acceleration in the field be a. Calculate where it should end up on the screen in terms of a.
 
WestTXcollegekid said:
The issue I'm having is, how do i go about finding the y velocity after it exits the E field. and how do I find the acceleration on the object by the E field if i don't know the displacement during its time in the E field.
Since these things depend upon quantities you don't know (the mass and charge), you should leave them as variables.

Edit: Note that the problem wants you to find the ratio q/M. Figure that at some point you'll be able to insert some variable (maybe call it "r") to represent q/M. I'd take a look at your acceleration and see what it depends on.
 
A small object with mass m, charge q, and initial speed v0 = 6.00×103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1) . The electric field between the plates is directed downward and has magnitudeE = 700 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distanced = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance. Calculate the object's charge-to-mass ratio, q/m. can anyone help me with this ques please
 
ur sh,

If you want help with the same problem, then you'll have to supply your own attempt at a solution first. The Original Poster in this thread still has not solved the problem, so getting him/her to the finish line will take priority if he/she returns.

Note that according to the forum rules, no help can be offered until an attempt at solution has been made.
 
  • #10
A small object with mass m, charge q, and initial speed v0 = 6.00×103 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm (Figure 1) . The electric field between the plates is directed downward and has magnitudeE = 700 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distanced = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.
Calculate the object's charge-to-mass ratio, q/m.
but this ans is wrong
E=F/q
700=F/q
700q*82=1/2mv^2
2*700*82/(6*10^3)=(1400*82)?36*10^6
32*10^-4
 
  • #11
You'll have to explain your logic. What does your third step represent?
 
  • #12
gneill said:
You'll have to explain your logic. What does your third step represent?

Hmm, how does he get 82?
 
  • #13
82 is the total distance traveled by the particle
 

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