- #1

MissPenguins

- 58

- 0

## Homework Statement

A charged mass on the end of a light string is

attached to a point on a uniformly charged

vertical sheet (with areal charge density

0.25 μC/m

^{2}) of infinite extent.

Look at attachment for diagram

Find the angle [itex]\theta[/itex] the thread makes with the

vertically charged sheet. The acceleration of

gravity is 9.8 m/s

^{2}and the permittivity of free

space is 8.854 × 10

^{−12}C

^{2}/N · m

^{2}.

Answer in units of degree.

I got the first part right to be 21.254189 using [itex]\theta[/itex]=tan

^{-1}(qE/mg)

I need help with part 2

Find [itex]\sigma[/itex] for an angle of 77

^{o}.

Answer in units of μC/m

^{2}.

## Homework Equations

[itex]\theta[/itex]=tan

^{-1}(qE/mg)

qE = [itex]\sigma[/itex]/2E

_{o}

## The Attempt at a Solution

Since [itex]\theta[/itex]=tan

^{-1}(qE/mg), so I did tan(77

^{o}) = (qE)(mg)

qE = 0.0038118 (calculated from part 1)

mg = 0.0098

[itex]\sigma[/itex]=(qE)(mg) = 7.51677X10

^{-13}

Converted back to uC = 7.51677X10

^{-7}

**What did I do wrong? Thank you!**