# Several Electrostatic Problems

1. Sep 1, 2006

### mrjeffy321

1) At the two ends of a pole there are two + charged spheres…one has a charge n times that of the other (nq and q) [I actually have a numerical value for n to use to find my final answer]. Along this pole, of length we will call d, there is another charged sphere which is located at an equilibrium position we will call x units away from the first sphere. My task is to find the distance x. I will call the distance of the middle sphere from the “right” sphere y = d – x.

To solve this one needs to realize that in order for the middle sphere to be at equilibrium the net force acting on it must be zero. There the force acting on it from the “left” sphere must equal the force acting on it from the “right” sphere.
Setting the two electrostatic forces equal to each other, I get,
n / x^2 = 1/y^2 --> y * sqrt (n) = x
(d – x)*sqrt (n) = x --> x = d * [sqrt (n) / (1 + sqrt(n))]
When I plug in my numerical value for n, I get an answer which seems reasonable and checks out mathematically by choosing some arbitrary value of d, but apparently it is wrong (the computerized homework system doesn’t accept it).
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2) In this problem, two charged spheres hand from the ceiling on strings (of negligible mass of course…this is a physics problem). Each sphere has a mass of m grams, has a charge of +c C on one and -c C on the other, and hangs at an angle of θ degrees away from the vertical. There is some “extra” electric field applied to the space around the spheres, call it E, which I am trying to find.
I have actual, numerical, values for m, c, and θ to use for calculation purposes.

To solve this problem, I found the weight of each sphere, the found the electrostatic force required for each sphere to hand at the angle that do (resultant force acting on sphere from gravity and electric forces must be equal in magnitude to the force of tension in the string, but in the opposite direction. Drawing the force diagram, the force of weight, mg, is direction downward and the electrostatic force for both spheres are directed away from the vertical.
Once I found the electrostatic force I was able to find the net electric field at that point (F_e = q * E). I found the electric field to be about ___ kN/C (the actual value is irrelevant, assuming I calculated it correctly). But that is the “net” field (if there was such a word). Remembering that since the spheres are oppositely charged, there is some attractive force between them which needed to be overcome by this extra applied field. I found what the electric field would be had there not been this extra applied field and if the two spheres were in their current position, this field was much stronger, over 10 times stronger, and in the “opposite” direction. So to solve for the applied field, I simply subtracted the “net” field from what the field would have been without it and found the difference (about ____ kN/C).
But apparently this is wrong too and I not quite as confident in this answer as I am in the first question. In solving this problem, I totally ignored the fact that there were two charged spheres until the very end, should I have done this?
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3) In the next problem there is a thin charged wire which is bent into a semicircle.
The wire has a linear charge density of lambda = lambda_0 * cos (θ), where θ is the reference angle as measured from the straight up vertical.
At the center of what would be the circle (at radius R away from all points along the wire), sits a point charge of magnitude c C.
I am at a loss for how to solve this problem. Conceptually, I know that the force will be entirely directed downward and away from the wire since the sides of the semi-circle cancel each other out in the +- X direction, for the only force in the –Y direction. However, I am not sure how to figure out just what this force is.
The charge on the wire is symmetrical along the wire around the “top” of the semi-circle, lessening nearer the edges.

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I could provide more numbers of pictures if it would help.
On the first two questions, I think I have the procedure down (maybe), but the numbers just aren’t coming out…on the 3rd question, I don’t really know where to begin.

2. Sep 1, 2006

### Saketh

What are you looking for in this one?

To find the electric field of the semicircle:
$$E = \int{dE} = \int{\frac{dq}{4\pi \epsilon_0 r^2} \hat{r}}$$.

You also know that $$dq = \lambda \cos{\theta} R d\theta$$. Remember, when you integrate, to keep the cosine inside the integral.

3. Sep 1, 2006

### mrjeffy321

The goal on question 3) is to find the total force acting on the charged object at the center of the charged semi-circle.

In the Integral you have to find the electric field, everything in it, but theta, is a constant, so it just comes down to integrating cos(theta) which is not hard. However, in order to actually use it in my calculation, would it not need to be a definite integral, integrating from -pi/2 to pi/2?
So it would be,
(K_e * lambda_0 / R) * [sin(theta)] evaluated at theta = -pi/2 to pi/2,
giving me,
-2 * K_e * lamba_0 / R, all of which are constants. If this is the electric field, then the rest of the problem is trivial.
[Ah , but no, the numbers still arent coming out right, so says the computer]

4. Sep 2, 2006

### mrjeffy321

OK, I figured out numbers 1) and 2)...you wouldnt belive what was wrong, it was due to the fact that I did not put a leading zero on my answer when I was entering it into the computer (the correct format was 0.123 rather than just .123).

But I am still having problems in the semi-circle question, as well as with a new question....

4) In this question, an infinitely long wire with a uniform linear charge density of lambda_0 is placed along a coordinate grid. The wire starts at a point given as -y_0 and extends on in the -y direction forever.
I am asked to find the electric field at the origin.
This is a purely symbolic problem, no numbers.

I think I should integrate and sum up the charge on the wire.
I think it should be,
Integral of ( ((K_e * lambda_0) / r) dr) from -y_0 to negative infinity

Does this look right?

Last edited: Sep 2, 2006