A grain of dust with a charge of -5 * 10 ^ -9 C is in equilibrium in a uniform electric field of a flat capacitor

A grain of dust with a charge of -5 * 10 ^ -9 C is in equilibrium in a uniform electric field of a flat capacitor with a strength of 6000 V / m if the field lines of the field are directed vertically downward. a) Find the mass of a grain of dust b) what should be the voltage between the plates, located at a distance of 5 cm from each other, in order to keep a grain of dust with a mass 2 times larger? c) Determine what charge the speck of dust has after engagement if it began to move with an acceleration of 8 m / s ^ 2. How many electrons did the speck lose?

a) The force acting from the electric field Fe = qE (directed upwards) balances the force of gravity Ft = mg.
qE = mg (1);
m = qE / g (2),
where m is the mass,
q – charge (absolute),
E – field strength,
g is the acceleration of gravity.
m = (5 * 10⁻⁹ Kl * 6000 V / m) / 9.8 m / s² = 3.06 * 10⁻⁶ kg.
Answer: The mass of a dust grain is 3.06 * 10⁻⁶ kg.