physics

Question

An oil droplet with a mass of 3.50 x 10^-15 kg accelerates downward at a rate of 2.50 m/s² when it is between two horizontal charged plates that are 1.00 cm apart. Assuming that the excess charge on the droplet is negative and the top plate is positive:

a) If the potential difference between the plates is 533 V, the charge on the droplet is ____ x 10^-19 C 

(give your answer in correct significant digits and do not include units)

b) How many excess electrons does the droplet carry?

Answer 1

(a)

Mass of the droplet is 3.50 x 10^-15 kg .

Downward acceleration of the droplet is a = 2.50 m/s^2 .

Distance between plates d = 1 cm = 1 × 10^-2 = 0.01 m.

Potential difference between the plates v = 533 .

The electric field intensity E = v/d

Substitute : d = 0.01 , v = 533 .

E = 533 / 0.01 = 533000 = 5.33 × 10^{4 .}

Net downward force = m × (downward acceleration) 

Since "mg" is pulling down and "qE" is pulling up, the net downward force is "mg−qE": 

mg−qE = m × (downward acceleration)

mg−qE = m × a

qE = mg - ma

qE = m [g - a ]

q = m [g - a ] / E 

q = (3.50 x 10^-15) [ 9.8 - 2.5 ] /  5.33 × 10⁴ 

q = (3.50 x 10^-15) [ 7.3  ] /  5.33 × 10⁴ 

q = 4.7936 × 10^-19 .

Therefore the charge of oil drop is  4.7936 ×  10^-19 C .

Answer 2

(b)

The charge of oil drop is q = 4.7936 × 10^-19  C .             [From (a)]

To find out the number of excess electrons of droplet, divide the charge of droplet with charge of electron .

Number of excess electrons = (charge of droplet) / ( charge of electron )

                                          = (4.7936 × 10^-19) / ( 1.60217657 × 10^-19 coulombs )

                                          = 2.9919 ≈ 3 .

Number of excess electrons = 3 .