What is the magnitude of qsoot (in units of c) so that the electric force exerted on the particle is equal to its weight = mg ? .
use: g = 10 m/s², and k = 9e9 n m²/c².
how many electrons must be added to the soot particle ?

In the exercise there are lost data, but I already found the exercise in a book

qsoot= 6.66e-18

no. electrons = approximately 42 electrons

Explanation:

msoot = 1.2e-12 kg

qcollector = 2e-06 C

rcollector = 0.1 m

The electric force must be equal to the weigth, so:

W=msoot*g                                          (1)

Felec = k*qsoot*qcollector/r^2           (2)

equaling (1) and (2):

msoot*g = k*qsoot*qcollector/r^2      (3)

reorganizing the equation (3) and clear qsoot:

qsoot = (msoot*g*r^2)/(k*qcollector)

qsoot= 6.66e-18

no. electrons =  qsoot / qelectron

no electrons = 41.6 electros (approximately 42 electrons)

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each  column  of the  table  is called a group. elements in a group share similar properties. groups are  read  from  top  to  bottom. metal. metalloid. nonmetal. solid. liquid. gas. fe. hg. o.  2. 3. 4. 5. 6. 7. 8. 9.  2. beryllium. 9.012. 4.  sodium. 22.990.  11. 3.  magnesium. 24.305.  12.  potassium. 39.098.  19. 4.  calcium. 40.078.  20. i guess  i