Question

Suppose a 2.0×10−62.0×10−6-kgkg dust particle with charge −1.0×10−9C−1.0×10−9C is moving vertically up a chimney at speed 6.0 m/sm/s when it enters the +2000-N/CN/C E⃗ E→ field pointing away from a metal collection plate of an electrostatic precipitator. The particle is 4.0 cmcm from the plate at that instant. Find the time needed for the particle to hit the plate. Express your answer with the

Answer 1

Answer:

Explanation:

mass of particle m = 2 x 10⁻⁶ kg

charge q = 1 x 10⁻⁹ C

electric field E = 2000 N/C

force on charge = E q

= 2000 x  1 x 10⁻⁹

acceleration = force / mass

= 2000x10⁻⁹ / 2 x 10⁻⁶

= 1 m /s²

initil velocity u = 6 m /s

distance s = 4 x 10⁻²

time = t

s = ut + .5  t²

4 x 10⁻² = 6t + .5 x 1 x t²

t² + 12t - .08 = 0

= .0067 s .

= 6.7 ms .