# Determine the speed of the particle when the string is parallel to the electric field.

A particle having charge q = +1.90 µC and mass m = 0.010 0 kg is connected to a string that is L = 1.60 m long and tied to the pivot point P in the figure below. The particle, string, and pivot point all lie on a frictionless, horizontal table. The particle is released from rest when the string makes an angle θ = 60.0° with a uniform electric field of magnitude E = 280 V/m. Determine the speed of the particle when the string is parallel to the electric field.
m/s

asked Jan 20, 2012 in Physics

+1 vote

Electric Potential of a Uniform Field is:

V = -E*ds, when you take V=0 at point P then the potential at the original position of the charge is  -E*s = -E*L*cos(theta), and at the final position you have V = -E*L (because cos(0) = 1).

Now use conservation of energy to solve:

Ki + Ui = Kf + Uf (K=kinetic, U=potential, i=initial, f=final)

plug in the the components.

0 - q*E*Lcos(theta) = (1/2)*m*v^2 - q*E*L

where q = 1.9*10^-6 C  (remember to convert from micrometers to meter)

E = 280 V/m

L = 1.6 m

m = 0.010 kg

theta = 60

Now plug in the variables and rearrange the equation solving for v:

v = squareroot ((2*(-q*E*L*cos(theta) + q*E*L) / m))

v = squareroot ((2*(-1.9*10^-6 C*280*1.6*cos(60) + 1.9*10^-6 C*280*1.6 m) / 0.010 kg))

and so

v  = 0.292 m/s
answered Jan 24, 2012 by ~Expert~ (1,216 points)
selected Jan 24, 2012 by Roar3215