How to find the magnitude of the electric field?

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SUMMARY

The discussion focuses on calculating the magnitude and direction of the electric field acting on a charged ball with a mass of 0.0024 kg, which falls with an acceleration of 3.13 m/s². The charge on the ball is determined to be 1.6E-8 C, derived from adding 1E11 electrons. The net force acting on the ball is calculated using Newton's second law, leading to the conclusion that the electric field strength is approximately 1E6 N/C directed downward.

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  • Understanding of Newton's second law (F = ma)
  • Knowledge of electric charge calculations (q = Ne)
  • Familiarity with free-body diagrams (FBD)
  • Concept of electric fields (E = F/q)
NEXT STEPS
  • Study the principles of electric fields and forces in physics.
  • Learn how to draw and interpret free-body diagrams for various scenarios.
  • Explore the relationship between mass, weight, and acceleration in gravitational fields.
  • Investigate the effects of electric fields on charged particles in motion.
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Students studying physics, particularly those focusing on electromagnetism and mechanics, as well as educators seeking to clarify concepts related to electric fields and forces.

Apollinaria
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I will keep working on this while waiting for someone here to respond...

Homework Statement



A ball m= 0.0024kg is electrically charged when 1E11 electrons are added. The ball falls with a=3.13m/s2 in an electric field. Find magnitude and direction of field.

Homework Equations



q= Ne

The Attempt at a Solution



If the ball is falling in the elec. field, it means it has a positive charge. If it's falling, it's going down so I have my direction for the elec. field.

I'm guessing that the charge on it is the number of electrons added * charge of electron.
So, 1E11 x (1.60E-19)= 1.6E-8 charge on ball.

F= qE. I don't have E. E=kq/r2, I don't have r.

And now?... :rolleyes:
 
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Electrons have a negative charge. So what sign is the charge on the ball? What are the units of charge?

What other force is acting on a ball of mass 0.0024 kg when it falls? What's the usual acceleration when a mass falls? How does it compare to the observed acceleration? What does that tell you about the force on the ball due to the electric field (direction, size)?
 
gneill said:
Electrons have a negative charge. So what sign is the charge on the ball? What are the units of charge?

What other force is acting on a ball of mass 0.0024 kg when it falls? What's the usual acceleration when a mass falls? How does it compare to the observed acceleration? What does that tell you about the force on the ball due to the electric field (direction, size)?

I know they do. Based on that the charge should be negative. However, if the ball is falling, it means its going where a positive charge would go in that field. So it must be positive. This is based on what we were taught :/

Gravity is acting on it at 9.8m/s2 but, I thought the acceleration already included gravity and was 3.13m/s2 TOTAL. As in, no need to include gravity.
Force would be 9.8 x 0.0024kg= 0.02352N
 
Apollinaria said:
I know they do. Based on that the charge should be negative. However, if the ball is falling, it means its going where a positive charge would go in that field. So it must be positive. This is based on what we were taught :/

Gravity is acting on it at 9.8m/s2 but, I thought the acceleration already included gravity and was 3.13m/s2 TOTAL. As in, no need to include gravity.
Force would be 9.8 x 0.0024kg= 0.02352N

The acceleration observed is the NET acceleration due to the sum of all forces acting. So rather than falling at g = 9.81 m/s2 which would be the rate if gravity alone were acting, the ball is observed to be falling at a = 3.13m/s2. Find the "missing" force.

Once you find the "missing" force, determine what electric field would be required to cause it on the given charge.
 
gneill said:
The acceleration observed is the NET acceleration due to the sum of all forces acting. So rather than falling at g = 9.81 m/s2 which would be the rate if gravity alone were acting, the ball is observed to be falling at a = 3.13m/s2. Find the "missing" force.

Once you find the "missing" force, determine what electric field would be required to cause it on the given charge.

Sigh... So I need to find the difference between the two of them?
9.81-3.13= 6.68.
I really don't understand yet.
 
Why don't you start by finding the net force which would produce the observed acceleration. Then draw a free-body diagram (FBD) for the ball showing the forces acting. Solve for the "missing" force.
 
gneill said:
Why don't you start by finding the net force which would produce the observed acceleration. Then draw a free-body diagram (FBD) for the ball showing the forces acting. Solve for the "missing" force.

The problem is that I'm unsure of what I'm supposed to be drawing. I don't know if my FBD is correct. I have the object, with mg pulling down. And F also pulling down? Or is it going the opposite way? I don't understand what the missing force is.
 
Apollinaria said:
The problem is that I'm unsure of what I'm supposed to be drawing. I don't know if my FBD is correct. I have the object, with mg pulling down. And F also pulling down? Or is it going the opposite way? I don't understand what the missing force is.

If the object is falling with an acceleration less than usual, which direction do you think the "missing" force is acting?

The "missing" force is the electric force on the charged ball due to the electric field acting on its charge.
 
gneill said:
If the object is falling with an acceleration less than usual, which direction do you think the "missing" force is acting?

The "missing" force is the electric force on the charged ball due to the electric field acting on its charge.

Its acting in the direction opposite of the mg. But how do I find the force if all I have is the different accelerations?
FE= ma, 0.0024kg x 6.68 = 0.016032N
W=mg, 0.0024kg x 9.81 = 0.023544N

ma-mg= (-)0.007512 N total difference.
 
  • #10
Apollinaria said:
Its acting in the direction opposite of the mg. But how do I find the force if all I have is the different accelerations?
FE= ma, 0.0024kg x 6.68 = 0.016032N
W=mg, 0.0024kg x 9.81 = 0.023544N

ma-mg= (-)0.007512 N total difference.

You're dancing all around the solution but somehow not managing to step on it :smile:

If you know the net (observed) acceleration (3.13 m/s2) and the mass, then you know the net force acting via Newton's 2nd law: fnet = m*a.

This net force is comprised of the sum of two forces (fG, fE) acting in opposite directions. So fnet = fG - fE. Solve for fE.

Now you have fE, which is directed upwards. It's produced by some electric field E acting on the charge on the ball...
 
  • #11
It's panic... I did what you recommended but it's not the right answer.
Answer is 1E6 N/C down.

Fnet= ma= 0.007512 N
Fnet= FE-Fg
0.007512 N + Fg = FE
0.007512 N + mg = FE
0.007512 N + (0.0024kg x 9.81) =0.031056 N upward direction.

F=qE
F/q=E
0.031056 N / 1.6E-8 = 1.94E6 N/C (down?)
 
  • #12
Apollinaria said:
It's panic... I did what you recommended but it's not the right answer.
Answer is 1E6 N/C down.

Fnet= ma= 0.007512 N
Fnet= FE-Fg
Okay, there's a small problem with choice coordinate axes and directions. If the ball is falling and you take that as the "positive" direction for the axis, then gravitational force will be in the positive direction, while the retarding electrical force will be in the negative direction.

Your Fnet above is a positive value, so it points down in the assumed "positive" direction of the axis. That would make Fnet = Fg - FE, since FE would be upwards for this choice of axis direction.

You should arrive at FE = 0.016N upwards.
 
  • #13
gneill said:
Okay, there's a small problem with choice coordinate axes and directions. If the ball is falling and you take that as the "positive" direction for the axis, then gravitational force will be in the positive direction, while the retarding electrical force will be in the negative direction.

Your Fnet above is a positive value, so it points down in the assumed "positive" direction of the axis. That would make Fnet = Fg - FE, since FE would be upwards for this choice of axis direction.

You should arrive at FE = 0.016N upwards.

You are a saint. It's been a long frickin day. I've been at this for the past 7 hours.
Thank you and I am forever grateful for all of your explanations and for your seemingly endless patience. Oh me.. :rolleyes:
 
  • #14
Apollinaria said:
You are a saint. It's been a long frickin day. I've been at this for the past 7 hours.
Thank you and I am forever grateful for all of your explanations and for your seemingly endless patience. Oh me.. :rolleyes:

No worries. Glad to help!
 

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