Electricity field, finding charge

In summary, the problem involves a charged particle of mass m suspended on a massless string in the presence of a uniform electric field. The electric field is given by the equation \vec{E} = \lambda \hat{i} + \mu \hat{j} and the particle is in equilibrium at an angle \theta, where \lambda and \mu are both positive. The task is to determine the charge q on the particle in terms of m, g, \lambda, \mu, and \theta. After setting up the coordinate system and identifying the main forces acting on the particle (gravity and the electric field), the solution involves solving for q using the equations F_{e} + F_{g} + F_{T} =
  • #1
cogman
5
0

Homework Statement


A charged particle of mass m is suspended on a massless-string in the presence of a uniform electric field. When the electric field is [tex]\vec{E} = \lambda \hat{i} + \mu \hat{j} N/C[/tex] the ball is in equilibrium at [tex]\angle \theta[/tex] ([tex]\lambda > 0[/tex] and [tex] \mu > 0[/tex])

Determine the charge q on the object in terms of m, g, [tex]\lambda, \mu, and \theta[/tex]


Homework Equations


[tex]F_{g} = m * g[/tex]
[tex]F_{e} = \vec{E} * q[/tex]


The Attempt at a Solution


I'm not to confident in what I have (hence the reason I'm posting here) but I believe it looks something like this.

First I would set up my coordinate system such that the x-axis is perpendicular to the string and the y-axis is parallel. With that, I could say that the two main forces acting on the particle are that of gravity and the electric field. Thus

[tex]F_{g} = F_{e}[/tex]
[tex]mg * (cos(\theta)\hat{i} + sin(\theta)\hat{j}) = E * q[/tex]
[tex]mg * (cos(\theta)\hat{i} + sin(\theta)\hat{j}) = (cos(\theta)(\lambda + \mu)\hat{i} + sin(\theta)(\lambda + \mu)\hat{j})q[/tex]

And that is about as far as I get before I start panicking. The major issue is, you can't divide vectors, so this is obviously has some serious flaw to it.
 
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  • #2
The electric field
[tex]\vec{E} = \lambda \hat{i} + \mu \hat{j}[/tex]
is a given quantity, so all parts of it are specified as part of the problem - including the unit vectors [itex]\hat{i}[/itex] and [itex]\hat{j}[/itex]. The point is, since those are given, you can't just arbitrarily decide that, say, [itex]\hat{j}[/itex] points along the string. You have to use the coordinate system you're given. (Well, you could always rotate the whole problem into another coordinate system, but that would be completely unnecessary here.)
 
  • #3
Ok, taking your advice (sorry for the delay in posting) here is my second stab at it.

[tex]F_{e} + F_{g} + F_{T} = 0[/tex]
[tex]F_{e} = q\lambda \hat{i} + q\mu \hat{j}[/tex]
[tex]F_{g} = -mg \hat{j}[/tex]

We Should be able to say that

[tex]F_{T}\hat{i} = -q\lambda \hat{i}[/tex]
because gravity doesn't contribute anything in the [tex]\hat{i}[/tex] direction. For the [tex]\hat{j}[/tex] we would say
[tex]F_{T}\hat{j} = -q\mu \hat{j} + mg\hat{j}[/tex]

With this in mind, we can bring up the formula that

[tex]|F_{T}|*sin(\theta) = |F_{g}\hat{j} + F_{e}\hat{j}|[/tex]

For this, we need to find [tex]|F_{T}|[/tex], given our previous definition, that is simply

[tex]\sqrt{(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}}[/tex]

thus

[tex]\sqrt{(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}} * sin(\theta) = \sqrt{(mg)^{2} + (q\lambda)^{2}}[/tex]

From there, it is just algebra to solve for q. Squaring everything, I get

[tex](q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}) * sin^{2}(\theta) = (mg)^{2} + (q\lambda)^{2}[/tex]

[tex](q\lambda)^{2} + (q\mu)^{2}) * sin^{2}(\theta) - (q\lambda)^{2} = (mg)^{2} - (mg)^{2} * sin^{2}(\theta)[/tex]

[tex]q^{2}(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2}) = (mg)^{2} - (mg)^{2} * sin^{2}(\theta)[/tex]

[tex]q^{2} = \frac{((mg)^{2} - (mg)^{2} * sin^{2}(\theta))}{(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2})}[/tex]

so for a final answer it would look something like

[tex]q = \sqrt{\frac{((mg)^{2} - (mg)^{2} * sin^{2}(\theta))}{(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2})}}[/tex]

noting that there is no longer a - charge as the problem stated that mu and lambda are always positive.

Did I do the math right and make the correct assumptions?
 
  • #4
see the papers of myron evans (alpha inst. for adv. lrn.)
 
  • #5
vitruvianman said:
see the papers of myron evans (alpha inst. for adv. lrn.)

Thanks, I'll probably check it out later, however, Theoretical physics is probably not what my teacher wants to see in an introductory Electro-magnetic physics course.

I'm mostly just trying to confirm that I've done the math correctly (and if not, where did I go wrong)
 
  • #6
Turns out I'm an idiot. :)

long story short, I was trying to take the magnitude of a vector in the i and j direction by trying to square each component of that i and j vector. You can't do that. the final answer comes to be something like

[tex]|F_{T}|sin(\theta) = q\lambda[/tex]

[tex]|F_{T}|cos(\theta) = mg - q\mu[/tex]

[tex]q\lambda cot(\theta) = mg - q\mu[/tex]

[tex]q = \frac{mg}{\lambda cot(\theta) + \mu}[/tex]
 
  • #7
lloll
 

What is an electric field?

An electric field is a region in which electrically charged particles experience a force. It is created by electrically charged objects and can be represented by electric field lines.

How do you calculate the strength of an electric field?

The strength of an electric field is calculated by dividing the force exerted on a test charge by the magnitude of the test charge. The formula is E = F/q, where E is the electric field strength, F is the force, and q is the test charge.

What is electric charge?

Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field. There are two types of electric charge: positive and negative.

How do you find the charge of an object?

The charge of an object can be found by using Coulomb's law, which states that the force between two charged objects is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. By measuring the force and distance between two objects, the charge can be calculated.

What is the unit of electric charge?

The unit of electric charge is the coulomb (C). It is defined as the amount of charge that passes through a point in an electric circuit in one second when a current of one ampere is flowing.

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