# Charged mass on a string

1. Feb 6, 2006

### Punchlinegirl

A charged mass on the end of a light string is attached to a point on a uniformly charged vertical sheet of infinite extent. The acceleration of gravity is 9.8 m/s^2 and the permittivity of free space is 8.85 x 10^-12 C^2/Nm^2. Find the angle $$\theta$$ the thread makes with the vertically charge sheet. Answer in units of degrees.
Given:
mass of ball= 1 g
Areal charge density of the sheet= 0.23 $$\mu C/m^2$$
length of the string = 78.9 cm
Then force of charge= qE= q$$\sigma$$ / 2E_0
We did some of this problem in class and went through the long process of drawing a free body diagram and summing up the components, we found that it was easier to use the pythagorean theorem to solve for T.
I found that T= $$\sqrt (mg)^2 + (qE)^2$$
So T= $$\sqrt 96.04 + 1.32 x 10^-5$$
So T= 9.8.
Then I plugged it into what we got for the forces in the y-direction, which was $$\theta= cos^-1 (-mg/T)$$
So theta= cos ^-1 (-9.8/9.8)
= 180 degrees which is wrong... can someone help me please?

2. Feb 6, 2006

### Staff: Mentor

Check your calculation of $q E$ and $mg$ (note that m = 0.001 kg). What's $q$?

Last edited: Feb 6, 2006
3. Feb 6, 2006

### Punchlinegirl

Well I stupidly forgot to change to kg, but I'm still getting the wrong answer.
$$T= \sqrt (mg)^2 + (qE)^2$$
mg= .001 * 9.8 = .0098
qE= $$q \sigma/2 E_o$$
qE= 2.8 x 10^-7 * 2.3 x 10^-7 / 2 * 8.85 x 10^-12
qE= .00364
T= $$\sqrt (.0098)^2 + (.00364)^2$$
T= .0104
$$\theta= cos^-1 (-mg/T)$$
$$\theta = cos^-1 (-.0098/.0104)$$
$$\theta= 159 degrees$$

4. Feb 7, 2006

### Staff: Mentor

What's with the minus sign?
$$\theta= \cos^{-1} (mg/T)$$

Your calculation would be a bit easier if you used:
$$\theta = \tan^{-1} (qE/mg)$$
(This way you don't have to calculate T.)