Calculate Charge on Ball Given Electric Field and Mass

In summary: This is okay. But you can do this in one step: mg * r * tan α = q.but i am having trouble in solving for r; Why? You know all the other variables (m, g, α, and q). Just plug them in.i tried solving for r by using the Pythagorus theorem, but i didnt workout... I don't know what you're trying to do with the Pythagorean theorem. You don't need it for this problem.I'm kinda stuck, i guess its because the known value for E field is in a component form...No, that's not the problem.
  • #1
1517279
7
0

Homework Statement


An electric field E=141000.0i N/C causes the 4.70 g ball in the picture to hang at a α=22.9° angle. What is the charge on the ball?


Homework Equations


E=(1/4pieE)*(q/r^2)

The Attempt at a Solution


I tried solving this question first by drawing a force diagram on the charged particle. On the diagram:
W = T cos α -1
F = T sin α -2

I divided equation 2 by 1
I get tan α = F/mg ; since F = 1/4pie * q/r^2; i subbed it in
therefore: mgtanα4pie*r^2 = q

but i am having trouble in solving for r; i tried solving for r by using the Pythagorus theorem, but i didnt workout... I'm kinda stuck, i guess its because the known value for E field is in a component form...

can someone please guide me through this question



picture is attatched
thanks in advance.
 

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  • #2
What picture?
How long is the string?
 
  • #3
length of the string is not given
 
  • #4
1517279 said:

Homework Equations


E=(1/4pieE)*(q/r^2)
That's the field of a point charge--not relevant here. You are given the field.

The Attempt at a Solution


I tried solving this question first by drawing a force diagram on the charged particle. On the diagram:
W = T cos α -1
F = T sin α -2

I divided equation 2 by 1
I get tan α = F/mg ;
This is fine.
since F = 1/4pie * q/r^2;
This is not. What force does a field E exert on a charge q?
 
  • #5
k it should be f=1/4pieEo *q/r^2 then right?
 
  • #6
1517279 said:
k it should be f=1/4pieEo *q/r^2 then right?
No. Again, that's the field surrounding a point charge--not relevant here. You have a uniform field given; you need to find the force that given field exerts on the charge. (What's the definition of electric field?)
 

Related to Calculate Charge on Ball Given Electric Field and Mass

1. How do I calculate the charge on a ball given the electric field and mass?

To calculate the charge on a ball, you can use the formula Q = m x a, where Q is the charge, m is the mass, and a is the acceleration due to the electric field. First, determine the acceleration by dividing the electric field strength by the mass. Then, multiply the acceleration by the mass to find the charge.

2. What units should I use for the electric field and mass when calculating the charge?

When using the formula Q = m x a to calculate the charge, the electric field strength should be in units of Newtons per Coulomb (N/C) and the mass should be in units of kilograms (kg). This will ensure that the resulting charge is in units of Coulombs (C).

3. Can I use this formula for any type of ball?

Yes, this formula can be used for any ball as long as you have the necessary information about the electric field strength and the mass of the ball. The formula is not limited to a specific type of ball or material.

4. How does the electric field affect the charge on a ball?

The electric field is directly proportional to the acceleration of the charge. This means that the stronger the electric field, the greater the acceleration and therefore the greater the charge on the ball will be. A weaker electric field will result in a smaller charge on the ball.

5. What other factors may affect the charge on a ball?

Aside from the electric field and mass, the distance between the ball and the source of the electric field may also affect the charge. The closer the ball is to the source, the stronger the electric field will be and therefore the greater the charge on the ball will be. Additionally, the charge on the ball may also be affected by the material of the ball, as some materials may have a higher or lower charge capacity.

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