Study the diagram of Millikan's apparatus as shown in the figure, and use the information it contains to help you answer the questions that follow:

(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, F_e = 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.

(g) 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

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