Charged Point Mass in a Constant Electric Field

[rmunoz]

Homework Statement

A point particle with mass m and charge q is suspended from a fixed point by a massless thread of length\ell. This thread makes an angle of \theta with the vertical when a constant electric field of magnitude E is directed along an angle \phi above the horizontal direction. Derive an expression for the angle \theta assuming m,q, \ell, \theta, E, and \phi are known.

a) draw a picture of the situation described above
b) draw a fbd for the charged mass
c) identify each of the forces on the charged mass and write an expression for each of them in terms of variables listed above (where appropriate)
d) Using Newton's second law derive an expression for theta.

Homework Equations

F=ma

E= F/q => k(q/r^2) (point charge)

The Attempt at a Solution

So far i have drawn both the fbd and the picture describing the situation. What I need to know is whether or not I am representing the forces correctly in the FBD. Because I do not have a scanner I will attempt to explain how I have set up the FBD in my own words.

First there is a weight force acting downward on the charged particle... obviously in the negative \hat{j}.

Second there is a tension force acting at some angle theta from the horizontal... i represented this as T.

Third there is the Electric field acting on the particle at angle phi above the horizontal (we may assume this has some single magnitude and direction because the field is constant).

My primary question is how does the electric field differ from the force acting on the charged mass if at all? I understand that E= F/q_{o} for a point charge. I am confused as to weather the force acting on the particle in the FBD is represented as F or as E in this situation.

 

[rmunoz]

I believe i have made quite a bit of progress since the first post but perhaps someone could help by checking over my work.

c) *For Finding T with respect to the other quantities:

T= (mg - E(sin(phi))/(cos(theta))
and
T= (Ecos(phi))/(sin(theta))

* the magnitude of the force acting on the particle as a result of the uniform electric field is given by:

E= (1/(4pi(epsilon-not))) ( q/(r^2)) (r-hat)

d) finding an expression for theta using Newtons 2nd

f=ma => because this is an equilibrium situation ma=0 thus the summation of forces acting on the particle must also be = to zero

Summation of forces in the x = 0 yields:
T= (E(cos(phi)))/(sin(theta))

Summation of forces in the y = 0 yields:
T= (mg-E(sin(phi)))/(cos(theta))

combining the two equations yields:
theta = arctan((E(cos(phi))/((mg-(E(sin(phi)))