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Help electric charge Coulomb's law two charges?

  • Thread starter asdf12312
  • Start date
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1
1. Homework Statement
http://loncapa.vcu.edu/res/vcu/phys202/zzImages/two-charges-forces.bmp

Problem#1:
Two identical conducting balls, A and B, of identical masses m = 10 kg, are suspended in equilibrium by insulating massless strings in length L = 3 m. Both balls make the same angle θ = 30° with the vertical axis. Both masses have equal charge. You can ignore the size of the balls.

What is the magnitude of the Coulomb force, i.e. electric force, exerted on A from B due to the charges? (From free body diagram analysis)
What is the amount of charges (do not worry it is positive charge or negative charge) on each ball?



Problem#2:
Two identical conducting balls, A and B, of identical masses m = 40 kg, are suspended in equilibrium by insulating massless strings in length L = 2 m. Both balls make the same angle θ with the vertical axis. The angles are very small such that small angle approximation applies (i.e. tan θ = sin θ). Both masses have equal charge Q= 3e-06 C. You can ignore the size of the balls.

What is the distance r between the two balls?


2. Homework Equations
F=q1*q2*k/(r^2)
where k=9*10^9
E=F/Q


3. The Attempt at a Solution
the only part i figured out (sadly) is the 1st part to find the Coulomb force. and that was only because the teacher gave us the equation :(
((40kg*10)/cos(30))*sin(30)=i got 57.7N for coulomb force
still don't no why thats the answer, but at least i no how to get it.
but i don't know what to do from here. i don't know how to find r so i assume it is L given (=3m). so i don't know how to find q (amt of charges).

for problem#2 i am stumped. i don't understand angle approximation and i can't use the cos/sin method that our teacher gave us. so wat do i do??
 

tms

644
17
The problem gave you one small-angle approximation. Another is [itex]\sin\theta \approx \theta[/itex].

The first step to solve the problem is to draw a diagram showing the forces on each ball. From that calculate the horizontal forces. That equation will include [itex]r[/itex], which you can then solve for.
 
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T(y)=mg
F(t)=mg/cos(theta)
F(e)=mg*tan(theta)

do i need sin in one of the equations?? wat am i doing wrong. also i dont understand small angle approximation so how to find the angle if i know sin=tan?
 
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For problem #1, you can calculate r by recognising that the triangle made from the 2 strings and a line connecting the 2 charges is equilateral.
 

tms

644
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T(y)=mg
F(t)=mg/cos(theta)
F(e)=mg*tan(theta)
What are these supposed to be?
do i need sin in one of the equations?? wat am i doing wrong.
For one thing, you are not using conventional orthography; ignoring the conventions hinders communication.
also i dont understand small angle approximation so how to find the angle if i know sin=tan?
What don't you understand about the approximations? Look at their series expansions, and think about what happens to the sine and cosine when the angle is very small.
 
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What are these supposed to be?

For one thing, you are not using conventional orthography; ignoring the conventions hinders communication.

What don't you understand about the approximations? Look at their series expansions, and think about what happens to the sine and cosine when the angle is very small.
ohhh i see..yeah so now i kinda understand. cos approaches 1 at small angles, but i still dont know how to use the sin one.
so the F(t) tension force would be (40*10)/cos(0)=400N
but that means the total electric force is 0? because i multiply this by sin(0) and get 400*0=0.

i'm just approaching this like i got the answer to the 1st part on problem#1, coz the two seem very similar. but i have a feeling im doing something wrong.
 

tms

644
17
ohhh i see..yeah so now i kinda understand. cos approaches 1 at small angles, but i still dont know how to use the sin one.
Again, look at the series expansion of sine (remembering that the expansions use radians, not degrees). I don't know how to make it clearer.
so the F(t) tension force would be (40*10)/cos(0)=400N
That is the horizontal component of the tension. Except that the angle is not zero; it is small, but not zero. You will also need the vertical component,.
but that means the total electric force is 0? because i multiply this by sin(0) and get 400*0=0.
Not at all. Again, the angle is small, but not zero.
i'm just approaching this like i got the answer to the 1st part on problem#1, coz the two seem very similar. but i have a feeling im doing something wrong.
You will communicate better if you use conventional orthography and spelling. Things like 'kinda' and 'coz' are okay in casual speech, but writing is more formal, and such spellings and lack of capitals and so forth are at best distractions.
 
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series expansion?? all i know is that sin is like the inverse of cos, so sin(90) is same as cos(0). but i don't see how that helps me because i planned to use cos/tan like in the last one.

for the vertical component in my FBD i got mass*gravity=400N. is that right?

i still don't know how to find the very small angle so if i have to use cos(0.01) instead of cos(0).

Again, look at the series expansion of sine (remembering that the expansions use radians, not degrees). I don't know how to make it clearer.

That is the horizontal component of the tension. Except that the angle is not zero; it is small, but not zero. You will also need the vertical component,.

Not at all. Again, the angle is small, but not zero.

You will communicate better if you use conventional orthography and spelling. Things like 'kinda' and 'coz' are okay in casual speech, but writing is more formal, and such spellings and lack of capitals and so forth are at best distractions.
 

tms

644
17
The series expansion for sine is
[tex]
\begin{align}
\sin x &= \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!} \,x^{2n+1} \\
&= x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots.
\end{align}[/tex]
The other circular functions have similar expansions. In this case, you want to eliminate all terms in x to a higher power than 1.

The only vertical force is gravity. The tension has a vertical component, which must offset gravity. Since the angle is small ...

As for the angle, that is what you are solving for. isn't it?
 
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i plugged in 1 into the first 3 terms of the sine expansion and got 0.842. this is angle right, not radians?

for the coulomb force i got mass*gravity=(400N/cos(0.842))*sin(0.842)=5.88N

5.88=Q^2*k/r^2
r^2=(3e-06^2*9*10^9)/5.88
r^2=0.081/5.88
r=0.117m

but i'm still gettin the wrong answer. what did i do wrong?
 

tms

644
17
i plugged in 1 into the first 3 terms of the sine expansion and got 0.842. this is angle right, not radians?
Just use the first term of the expansion. That leads to the approximation [itex]\sin x = x[/itex], as I said above.
 
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still getting the wrong answer..don't know what i'm doing wrong.

400N*tan(1)=6.98

Q=3e-06 C
K=9x10^-9

6.98=Q^2*K^2/R^2

R^2=0.081/6.98
R=0.108m
 

tms

644
17
First, you can't just assume a particular value for theta. Use the approximations to find a relation between r and theta.
 
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sin(theta)=tan(theta)=theta=O/H=(r/2)/2m=r/4
400N*(r/4)=Q^2*K^2/R^2
400*(R^3)/4=0.081
R^3=0.00081
R=0.093

did i do this right?
 
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sorry O/H is opposite/hypotenuse. in the right triangle i got r/2 for the opposite (since r is line from A to B) and L=2m for hypotenuse. and no sorry i wasn't squaring k that was a typo.
 
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can you plz just tell me how to get the answer? i have tried but i just dont understand it o.o
 

tms

644
17
No; it is against the rules here to do that.

Let's start over from the beginning. Draw the diagram showing all the forces. Each ball is acted on by the tension in the string, gravity, and the electrostatic force. Since the system in in equilibrium, all the forces on each ball will add up to zero. You want to break the forces into their vertical and horizontal components (since the setup is symmetric, you only have to do this for one side.

So what are the components of the forces on the ball? You'll get two equations in two unknowns: [itex]T[/itex] and [itex]\theta[/itex]. Don't worry about the small-angle approximation yet.

Use symbols everywhere, not numbers, so you can see more easily what is going on. Let [itex]\theta[/itex] be the angle between the string and the vertically down direction.
 
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all i know is electrostatic/coulomb force=mass*gravity*tan(θ). and since the problem says tan(θ)=sin(θ), force=mass*gravity*sin(θ)

for the FBD..don't really know how to draw it. in the negative Y direction i get mass*gravity. don't know how to calculate string tension, i am assuming it is just L given.
 
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tms

644
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There are three forces on each charged body: gravity, tension, and electrostatic. Gravity is down, electrostatic is horizontal, tension is at an angle [itex]\theta[/itex] to the vertical. Since the bodies are at rest, the forces must cancel. That is, the vertical component of the tension must be equal and opposite to gravity, and the horizontal component of the tension must be equal and opposite to the electrostatic force.
 
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ok, so the vertical tension is mass*gravity=400N
the horizontal electrostatic force = KQ^2/R^2 = (9x10^9)(3e-06 C)^2/R^2 = 0.081/R^2
 

tms

644
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ok, so the vertical tension is mass*gravity=400N
That is the vertical component of the tension.
the horizontal electrostatic force = KQ^2/R^2 = (9x10^9)(3e-06 C)^2/R^2 = 0.081/R^2
You need to express the distance in terms of the angle and the length of the string.

It would be helpful to use symbols in your calculations until the very end. That way you can see what is going on better, and might find some simplifications that are hidden by the numbers.
 
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ok, so i have to find equation for R. length of string=2m.
how about R/2=sin(θ)*L
i am using pythagorean theorem. cut the triangle into two right traingles.
 

tms

644
17
Okay. Now set up the two equations for the horizontal and vertical forces. After you do that, use the small-angle approximations to eliminate [itex]theta[/itex].
 

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