In summary: Got it. Thanks!In summary, the problem involves three charged particles placed at the corners of an equilateral triangle. Two particles have negative charges while the third has a positive charge. The question asks for the net electric force acting on the positive particle due to the negative particles. Using the given equations and values, the components of the force acting on the positive particle are found to be (2.96x10^-4, 5.13x10^-4). The magnitude of the force is then calculated using the formula F = (Fx^2 + Fy^2)^1/2, resulting in a value of 1.19x10^-3. To find the angle between the force components, the inverse tangent function should
  • #1
whitejac
169
0

Homework Statement


Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Homework Equations


Fcos(θ) = Fx
Fsin(θ) = Fy
θ = Tanh(Fy/Fx)
F = (Fx^2 + Fy^2)^1/2
F(2on3) = K(q1q2)/r^2)

The Attempt at a Solution


Using the program's help function, I have found the components
(F1on3)x = 2.96x10^-4
(F1on3)y = 5.13x10^-4

So the only thing left was to find F(2on3) in the same way I found F(1on3).
Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this:

Fy = 0 (sin0=0)
Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2
= 1.19x10^-3

This makes sense given that it should attract.

Here's where the problem falls off of the rails...

I need to find Theta

So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006

I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?
 
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  • #2
whitejac said:

Homework Statement


Three charged particles are placed at each of three corners of an equilateral triangle whose sides are of length 2.7 cm . Two of the particles have a negative charge: q1 = -6.0 nC and q2 = -12.0 nC. The remaining particle has a positive charge, q3= 8.0 nC . What is the net electric force acting on particle 3 due to particle 1 and particle 2?

Homework Equations


Fcos(θ) = Fx
Fsin(θ) = Fy
θ = Tanh(Fy/Fx)
F = (Fx^2 + Fy^2)^1/2
F(2on3) = K(q1q2)/r^2)

The Attempt at a Solution


Using the program's help function, I have found the components
(F1on3)x = 2.96x10^-4
(F1on3)y = 5.13x10^-4

So the only thing left was to find F(2on3) in the same way I found F(1on3).
Arranging the problem so that q3 was at the origin, q1 was at the top of the triangle and q2 was along the x-axis, I found this:

Fy = 0 (sin0=0)
Fx = (9x10^9)(12x10^-9)(8x10^-9) / (2.7x10^-2)^2
= 1.19x10^-3

This makes sense given that it should attract.

Here's where the problem falls off of the rails...

I need to find Theta

So Tanh (5.13x10^-4)/((2.96x10^-4)+(1.19x10^-3)) = 0.006

I really need this to be in the vicinity of 19degrees. I do not understand what mistake I am making... Does anybody else?
You should be careful here with your notation. Tan (θ) is not the same function as tanh (θ).

https://en.wikipedia.org/wiki/Hyperbolic_function

If θ = tan-1 ((5.13x10-4)/((2.96x10-4)+(1.19x10-3)) is what you are trying to calculate ...
 
  • #3
Really? I thought i'd looked it up and found it was the same, math-wise, just different in definition. My mistake I guess...
Scientific calculators only offer the tanh function though. Is that because they expect you to simply find Tan(Θ) and then find 1/x?
 
  • #4
whitejac said:
Really? I thought i'd looked it up and found it was the same, math-wise, just different in definition. My mistake I guess...
Scientific calculators only offer the tanh function though. Is that because they expect you to simply find Tan(Θ) and then find 1/x?
I'm confused here.

The standard trigonometric functions are abbreviated as sin, cos, tan, cot, sec, and csc. cot = 1/tan; sec = 1/cos; and csc = 1/sin.

The inverse trig functions are sin-1, cos-1, etc.

Note that sin-1 ≠ 1 / sin. If y = sin (θ), then sin-1(y) = θ; -1 ≤ y ≤ 1

The hyperbolic functions sinh, cosh, tanh, coth, sech, and csch are defined:

exp_hyperbolic.gif

 
  • #5
Oh wow, yes I can see now how the hyperbolic functions are so much not the same thing...
Okay, I need to find tan^-1(Θ) in order to solve this problem and get me the angle between the force components? Sounds pretty straightforward now. I was making a mistake with my calculator because I didn't distinguish the notations.
 

Related to I'm making an Arithmetic Error, Electrostatic force diagrams

1. What is an arithmetic error in the context of electrostatic force diagrams?

An arithmetic error in electrostatic force diagrams refers to a mistake in the numerical calculations involved in determining the magnitude and direction of the electrostatic forces between charged particles.

2. How can arithmetic errors impact the accuracy of electrostatic force diagrams?

Arithmetic errors can greatly affect the accuracy of electrostatic force diagrams by leading to incorrect calculations of the forces between charged particles. This can result in misleading or incorrect conclusions about the behavior of these particles.

3. What are some common sources of arithmetic errors in electrostatic force diagrams?

Some common sources of arithmetic errors in electrostatic force diagrams include miscalculations, rounding errors, and incorrect use of mathematical formulas or equations.

4. How can scientists avoid making arithmetic errors when creating electrostatic force diagrams?

Scientists can avoid making arithmetic errors by double-checking their calculations, using precise and accurate mathematical formulas and equations, and using appropriate units and significant figures in their calculations.

5. What steps can be taken to correct an arithmetic error in an electrostatic force diagram?

If an arithmetic error is discovered in an electrostatic force diagram, the first step is to identify the source of the error. This could involve checking calculations, reviewing equations, or consulting with a colleague. Once the error is identified, it can be corrected by recalculating the forces using the correct values and methods.

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