Answer to Question #165830 in Physics for Emma Hemmerton

Question

The basis for separation of many materials, ranging from ores to recycled plastics, is by electrostatic charge. The materials are ground and charged. They are then allowed to fall through an electric field. Charged particles are deflected in the field and land in different hoppers.

The electric field is supplied by two plates 4 m long, held vertically, and 0.4 m apart. The potential difference between them is 60 kV.

a) If the ore particles gain a specific charge 1.6 × 10^-6 C kg^-1, find the acceleration of the particles in the horizontal direction.

b) Hence calculate the horizontal deflection of the charged particles when they reach the bottom of the plates. Give your answer to two significant figures.

Answer 1

(a) Let’s first find the electric field strength:

$$E=\dfrac{\Delta V}{d}.$$

Then, the electric force acting on the ore particles can be written as follows:

$$F_e=qE=\dfrac{q\Delta V}{d}.$$

Finally, from the Newton’s Second Law of Motion we can find the acceleration of the particles in the horizontal direction:

$$a=\dfrac{F_e}{m}=\dfrac{q\Delta V}{md}=\dfrac{q_{sp}\Delta V}{d},$$ $$a=\dfrac{1.6\cdot10^{-6}\ \dfrac{C}{kg}\cdot60\cdot10^{3}\ V}{0.4\ m}=0.24\ \dfrac{m}{s^2}.$$

b) Let's first calculate the time taken for the ore particles to fall between the plates:

$$y=\dfrac{1}{2}gt^2,$$ $$t=\sqrt{\dfrac{2y}{g}}=\sqrt{\dfrac{2\cdot4\ m}{9.8\ \dfrac{m}{s^2}}}=0.9\ s.$$

Finally, we can calculate the horizontal deflection of the charged particles when they reach the bottom of the plates:

$$x=\dfrac{1}{2}at^2=\dfrac{1}{2}\cdot0.24\ \dfrac{m}{s^2}\cdot(0.9\ s)^2=9.72\cdot10^{-2}\ m.$$