What Angle Does the Charged Mass Form with the Vertical Sheet?

[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?

 

[Doc Al]

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

 

[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

 

[Doc Al]

\theta= cos^-1 (-mg/T)

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.)