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.