Physics Problem 22: Mass of Charged Object in Electric Field

  • Thread starter Thread starter mustang
  • Start date Start date
  • Tags Tags
    Physics
AI Thread Summary
To determine the mass of a charged object floating in an electric field, the key equation is F = qE = mg, where F is the force, q is the charge, E is the electric field strength, m is the mass, and g is the acceleration due to gravity. Given a charge of 40 µC and an electric field of 572 N/C, the forces acting on the object must be equal for it to float. The weight of the object (mg) is balanced by the electric force (qE). By substituting the known values into the equation, the mass can be calculated. This problem illustrates the relationship between electric force and gravitational force in a uniform electric field.
mustang
Messages
169
Reaction score
0
Problem 22.
given: g=9.814m/s^2.
An object with a net charge of 40 uC is placed in a uniform electric field of 572 N/C, directed vertically.
What is the mass of this object if it floats in this electric field? In units of kg.
Note: What formula would you use?
 
Physics news on Phys.org
Use Equilibrium Condition, Relation b/w Field and Force
 
I need more help. What does "b" and "w" mean?
 
"b" and "w" don't mean anything. humanshu121 was using b/w to mean "between".
 
I having a hard time solving problem since I don't know what formula I should use and the values to "sub" in.
 
What is the formula for the force experienced by a charge q in an electric field of magnitude E?
 
Would that formula be Electric field strength=Coulomb constant* charge producing the field(q)/(distance)^2.
 
No -- that formula describes the strength of the field PRODUCED by a charge q.

You want the force EXPERIENCED by a charge q under the influence of a field from some other source.
 
Is the formula electric force=Coulomb constant*(charge1)(charge2)/(distance)^2
 
  • #10
Not exactly, but you could find the right one "hidden" in there (because in there you have included the expression that you wrote up above for the field produced by a charge).
 
  • #11
I'm comfused?
 
  • #12
You have two forces,

1.weight from the mass acting towardds the earth
2.an electric force from the electric field, the key equation here is that: the force on a charged particle is its charge times the field -> F = qE

Note: Coulomb law is that just mentioned without the charge of the particle in question. Coulmbs law is with q1 and q2 is between two paticles, in this case the electric field is probably produced by large metal plates with millions of particles in them - point is how the field is made is not important.
 
  • #13
If the field E1 produced by a point charge q1 is
E1 = keq1/r2

and the force exerted on point charge q2 by point charge q1 is
F12 = keq1q2/r2

just substitute the formula up above for E to get
F12 = q2E1

Here, you aren't dealing with 2 point charges. Instead you have 1 point charge in a uniform field from some unspecified source. But as dodger said, the formula above can be generalized, so the force on a charge q from any uniform field is
F = qE

Now you just have to find the conditions so that force exactly offsets the weight of the particle -- basically a geometry question.
 
  • #14
This subject should not be a confusing thing, a field is produced by charged stuff and this exerts a force on other charged stuff, the size of this force (as given by the equation i said) tells you how big a force acts on the charged thing. Stop and think about it - if you have a charged particle in an electric field and you double its charge the effect on it goes the same - the force on it doubles. The same effect would happen if the field was doubled - the force on it would double.
 
  • #15
Right.

And forget about my unnecessary "geometry" comment.

There's really no geometry to deal with here. One force (the weight) acts straight down and the other (the electrical force) acts straight up (do you see why?).

If the object floats, these two forces must be equal, right? Simple algebra. Just solve for m.
 
  • #16
So F=qE is bascially F=ma, Right.
 
  • #17
Right; well, F=qE=F=mg.

I assume you meant g, right?
 
  • #18
So the equation to solve this problem is qE=mg, where g=9.81m/s^2, E=572N/C, and q= 40*10^-6.
 
  • #19
Correct.
 
  • #20
Thanks!

Thank you!
 
Back
Top