(a) The upper plate in the diagram is connected to the
positive terminal of the supply. What does this tell you about the sign of the
charge on the droplet?
(b) What is the electric field strength between the two
plates?
(c) What is the weight of the droplet?
(d) What is the electric force acting on it when it is
stationary?
(e) What is the charge on the droplet? What is the
significance of this value?
(f) In Millikan's experiment, he included a source of β-radiation.
When an oil droplet was irradiated, it was suddenly observed to start moving
upwards. What explanation can you give for this?
(g) Assuming that the charge on the oil droplet has
increased because it had captured a single electron, what new value of the
voltage between the plates would you now expect 10 to hold it stationary?
Solution:
(a) Since the upper plate is positively charged, it attracts
negatively charged electrons and repels positively charged ions. Therefore, the
charge on the droplet must be negative.
(b) Electric Field Strength (E) = ?
Voltage (V) = 180V
Thickness (d) = 1cm = 10-2cm
We Know, E = $\frac{V}{d}$ = $\frac{{180}}{{{{10}^{ - 2}}}}$
= 18000V/m
The required electric field is 18 × 103V/m.
(c)
Given:
Mass(m) = 5.9 × 10-16Kg
Weight = mg = 5.9 × 10-16 × 10 = 5.9
×
10-15
(d) Given:
E = 18000V/m
Charge of electron = 1.6×10-19
Thus, Electric Force = eE
= 18000 × 1.6×10-19 = 2.88×10-15N
(e)
We Know, Fe =
mg
Or, qE = mg
Or, q = $\frac{{5.9 \times {{10}^{ - 16}} \times {\rm{
}}10}}{{18000}} = 3.2 \times {10^{ - 19}}$
It shows, the charge is 2 times the charge of electron.
(f) The beta radiation from the source knocked electrons off
the oil droplets, giving them a negative charge. The negatively charged oil
droplets were then attracted to the positively charged plate, causing them to
move upwards.
Initial Charge (q)=3.2×10-19C
According to the question.
Charge (q’) = 3.2×10-19C + 1.6×10-19=4.8×10-19
Now,
q’E = mg
or, q’ × $\frac{V}{d}$=mg
or, $\frac{q’V}{d}$=5.9
×
10-15
Solving for V by putting above values, we get
V = 122.91volt
Given:
Initial Charge (q)=3.2×10-19C
According to the question.
Charge (q’) = 3.2×10-19C + 1.6×10-19=4.8×10-19
Now,
q’E = mg
or, q’ × $\frac{V}{d}$=mg
or, $\frac{q’V}{d}$=5.9
×
10-15
Solving for V by putting above values, we get
V = 122.91volt