# Columb force question

1. Aug 29, 2006

### nateastle

a small .4g sphere carries a charge of 3 X 10^-10 and is attached to one end of a silk fiber 8cm long. The other end of the fiber is attached to a large vertical insulating sheet that has surface charge density equal to 25 X 10 -6 C/m^2. The sphere on the end of the fiber is repelled outward and finally reaches a state of equilibrium. Find the angle that the fiber makes with the vertical sheet.

Okay here is what I have done. I know I need to get the tan of theta. What I did is I calculated out the 2 forces. There is gravity which is Tcos(theta) = mg and there is the colomb force which is Tsin(Theta) which is qE. From there I am getting confused on what I should do next. Any help is much appreciated.

2. Aug 29, 2006

### mezarashi

Good start. You realize that this is a static force analysis problem.

Let's assume that the insulating sheet is on the right and the sphere is being repelled to the left. Breaking it down into x and y components:

x-axis:
- Electric force pushing to the left (Felectric)
- String tension pulling to the right (Fstring)

y-axis:
- Gravity pulling down (Fgrav)
- String tension pulling up (Fstring)

How many unknowns do we have and can we solve with the number of equations available? We know that we have two equations, one in x: Fnet_x = 0, and similiarly in y: Fnet_y = 0. There are two unknowns, the angle theta and the string tension T. You're on the right track ^^.

P.S. You'll also need to know that the electric field due to a large sheet charge is uniform everywhere and equal to $$E = \frac{\sigma}{2\epsilon_0}$$ and of course, the Colomb force is then simply Fe = qE as you have mentioned.

3. Aug 29, 2006

### Andrew Mason

$$T\cos\theta = mg$$

$$T\sin\theta = qE$$

$$tan\theta = \sin\theta/\cos\theta$$

Seems to me you have everything you need.

AM

4. Aug 29, 2006

### nateastle

okay so if I am correct I would take the atan of my answer and get about 6.2 degrees is this correct?

5. Aug 29, 2006

### Andrew Mason

You will have to show us how you got E.

AM

6. Aug 29, 2006

### nateastle

I got E by $$\sigma$$ = 25 * 10^-6 C/m^2 then I took that and divided by 2e0 with e0 being 8.85 *10^-12 I took that answer(1.4 * 10^6) and multiplied it by the charge of the ball which is 3 X 10^10 C which gave a force of 4.23 X 10^-4.