## charged mass on a string

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?
 Mentor Blog Entries: 1 Check your calculation of $q E$ and $mg$ (note that m = 0.001 kg). What's $q$?
 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 $$\theta= cos^-1 (-mg/T)$$ $$\theta = cos^-1 (-.0098/.0104)$$ $$\theta= 159 degrees$$

Mentor
Blog Entries: 1

## charged mass on a string

 Quote by Punchlinegirl $$\theta= cos^-1 (-mg/T)$$
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.)