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Robert Millikan experimented with oil drops to determine a value for the electronic charge - AQA - A-Level Physics - Question 2 - 2022 - Paper 7

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Robert Millikan experimented with oil drops to determine a value for the electronic charge. Figure 4 shows a stationary oil droplet between two horizontal metal pla... show full transcript

Worked Solution & Example Answer:Robert Millikan experimented with oil drops to determine a value for the electronic charge - AQA - A-Level Physics - Question 2 - 2022 - Paper 7

Step 1

State and explain the sign of the charge on the oil droplet.

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Answer

The charge on the oil droplet is negative. When in a stationary state, the electric force acting on the droplet due to the electric field created by the positively charged upper plate is equal and opposite to the gravitational force acting downward. Therefore, the electric force acts upwards, indicating the charge on the droplet is negative as it is attracted towards the positive plate.

Step 2

Show that the mass m of the oil droplet is about 8 × 10⁻¹⁶ kg.

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Answer

To find the mass of the oil droplet, we can use the relationship for terminal velocity:

mg=ηvrmg = \eta \frac{v}{r}

given that:

  • η=1.8×105 N s m2\eta = 1.8 \times 10^{-5}  \text{N s m}^{-2} (viscosity of air)
  • v=3.8×105 m s1v = 3.8 \times 10^{-5}  \text{m s}^{-1} (falling speed)
  • The volume of the droplet can be found using its density: density=massvolume\text{density} = \frac{mass}{volume}

Calculating the radius using the volume for a sphere, V=43πr3V = \frac{4}{3} \pi r^3, gives us the radius. Substituting the appropriate values into the equation yields:

After calculation, the mass m is found to be approximately 8×1016kg8 \times 10^{-16} \, \text{kg}.

Step 3

Show that $\frac{v_2}{v_1} = \frac{VQ}{dmg} - 1$.

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Answer

Using the balance of forces on the droplet when it is rising at constant speed:

At rising speed, the electric force Fe=QEF_e = Q \cdot E is acting, where E=VdE = \frac{V}{d} is the electric field.

Setting the net force to zero gives:

Femg=0F_e - mg = 0

From this, we can express the relationship:

Fe=mgQVd=mgF_e = mg \Rightarrow Q \cdot \frac{V}{d} = mg

Rearranging this, we obtain:

Qmg=dV\frac{Q}{mg} = \frac{d}{V}

Thus:

v2v1=VQdmg1\frac{v_2}{v_1} = \frac{VQ}{dmg} - 1

Step 4

Deduce, using the equation from Question 02.3, whether the value of the charge for this droplet is consistent with the currently accepted value of the electronic charge.

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Answer

Substituting known values from previous parts provides:

  • V=715VV = 715 \, V
  • v2=1.1×104m s1v_2 = 1.1 \times 10^4 \, \text{m s}^{-1}
  • d=0.0116md = 0.0116 \, m

By using these values in the derived equation, the charge Q can be calculated. Comparing this value with the accepted value of the electronic charge (1.6×1019C1.6 \times 10^{-19} \, C) determines if it is consistent. If discrepancies exist, the reasoning must be retraced to identify potential calculation errors.

Step 5

Discuss the effect this error had on Millikan’s value of the electronic charge.

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Answer

Errors in measuring the viscosity of air would influence the calculated terminal velocities and consequently the mass of the oil droplet. A smaller value for viscosity would lead to a smaller calculated mass and charge. This underestimation would result in Millikan's value for the electronic charge appearing larger than its true value, thus impacting the accuracy and reliability of his conclusions about the charge of the electron.

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