Homework Help: Electricity field, finding charge

1. Jan 8, 2010

cogman

1. The problem statement, all variables and given/known data
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 $$\vec{E} = \lambda \hat{i} + \mu \hat{j} N/C$$ the ball is in equilibrium at $$\angle \theta$$ ($$\lambda > 0$$ and $$\mu > 0$$)

Determine the charge q on the object in terms of m, g, $$\lambda, \mu, and \theta$$

2. Relevant equations
$$F_{g} = m * g$$
$$F_{e} = \vec{E} * q$$

3. 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

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

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.

2. Jan 8, 2010

diazona

The electric field
$$\vec{E} = \lambda \hat{i} + \mu \hat{j}$$
is a given quantity, so all parts of it are specified as part of the problem - including the unit vectors $\hat{i}$ and $\hat{j}$. The point is, since those are given, you can't just arbitrarily decide that, say, $\hat{j}$ 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. Jan 11, 2010

cogman

Ok, taking your advice (sorry for the delay in posting) here is my second stab at it.

$$F_{e} + F_{g} + F_{T} = 0$$
$$F_{e} = q\lambda \hat{i} + q\mu \hat{j}$$
$$F_{g} = -mg \hat{j}$$

We Should be able to say that

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

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

$$|F_{T}|*sin(\theta) = |F_{g}\hat{j} + F_{e}\hat{j}|$$

For this, we need to find $$|F_{T}|$$, given our previous definition, that is simply

$$\sqrt{(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}}$$

thus

$$\sqrt{(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}} * sin(\theta) = \sqrt{(mg)^{2} + (q\lambda)^{2}}$$

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

$$(q\lambda)^{2} + (q\mu)^{2} + (mg)^{2}) * sin^{2}(\theta) = (mg)^{2} + (q\lambda)^{2}$$

$$(q\lambda)^{2} + (q\mu)^{2}) * sin^{2}(\theta) - (q\lambda)^{2} = (mg)^{2} - (mg)^{2} * sin^{2}(\theta)$$

$$q^{2}(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2}) = (mg)^{2} - (mg)^{2} * sin^{2}(\theta)$$

$$q^{2} = \frac{((mg)^{2} - (mg)^{2} * sin^{2}(\theta))}{(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2})}$$

so for a final answer it would look something like

$$q = \sqrt{\frac{((mg)^{2} - (mg)^{2} * sin^{2}(\theta))}{(\lambda^{2} + \mu^{2}) * sin^{2}(\theta) - \lambda^{2})}}$$

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. Jan 11, 2010

vitruvianman

see the papers of myron evans (alpha inst. for adv. lrn.)

5. Jan 11, 2010

cogman

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. Jan 12, 2010

cogman

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

$$|F_{T}|sin(\theta) = q\lambda$$

$$|F_{T}|cos(\theta) = mg - q\mu$$

$$q\lambda cot(\theta) = mg - q\mu$$

$$q = \frac{mg}{\lambda cot(\theta) + \mu}$$

7. Jan 14, 2010

lloll