# Homework Help: Find an expression for the capacitor voltage

1. Feb 15, 2016

### neilson18

1. The problem statement, all variables and given/known data

An uncharged parallel-plate capacitor with spacing d is horizontal. A small bead with mass m and positive charge q is shot straight up from the bottom plate with speed v_0. It reaches maximum height y_max before falling back. Then the capacitor is charged with the bottom plate negative.

Find an expression for the capacitor voltage ΔVC for which the bead's maximum height is reduced to 1/2y_max. Ignore air resistance.

Express your answer in terms of the variables m, d, q, and appropriate constants.

2. Relevant equations
V = EΔs
E=f/q
f=ma
v_f^2 = v_o^2 + 2aΔx

3. The attempt at a solution

I'm not sure how to approach this problem. I thought you could use a kinematics equation to solve for acceleration and then use that to find force, then E field, then V, but I was getting an answer that depended on v_o which is not allowed.

We have:
v_i = v_o
v_f = 0
a = ?
t = ?
Δx = y_max = d

0 = v_o^2 + 2ad/2
=> a = -v_o^2/d
=> f = -mv_o^2/d
=> E = -mqv_o^2/d
=> V = -mqv_o^2

So now I'm not sure where to go from here.

2. Feb 15, 2016

### Staff: Mentor

For the first shot the only force acting on the bead is gravity, producing an acceleration of g downwards. What is the expression for $y_{max}$ under those conditions?

How would you have to modify the acceleration in order to reduce that height by half?