A student investigated the behaviour of small, positively charged oil drops in an electric field - Edexcel - A-Level Physics - Question 9 - 2023 - Paper 3

Using the formula for weight, the weight of the oil drop can be calculated as:

$W = mg$

where:

• Density, $ho = 920 ext{kg/m}^3$
• Volume, $V = \frac{4}{3} \pi r^3 = \frac{4}{3} \pi (1.78 \times 10^{-6})^3$

Calculating the volume gives us:

$V \approx 5.310 \times 10^{-18} m^3$

Now, substituting for mass:

$m = \rho V = 920 \times 5.310 \times 10^{-18} \approx 4.891 \times 10^{-15} kg$

Thus, the weight is:

$W = mg \approx 4.891 \times 10^{-15} kg \times 9.81 \approx 4.80 \times 10^{-14} N$

This weight corresponds to the electric force experienced by the oil drop:

$F = EQ = W$

Where $E$ is the electric field given by:

$E = \frac{V}{d} = \frac{4870}{0.0155} \approx 313548 N/m$

Finding $Q$:

$Q = \frac{W}{E} \approx \frac{4.80 \times 10^{-14}}{313548} \approx 1.53 \times 10^{-19} C$

Since this charge is not a whole number multiple of the charge of an electron ($e \approx 1.6 \times 10^{-19} C$), the expectation is not confirmed.

The student should wait a short while to ensure that the oil drop reaches its terminal velocity. Initially, the drop will be accelerating due to gravity until the drag force balances the weight. Waiting allows the drop to stabilize to a constant velocity, which gives accurate measurements.

To determine the terminal velocity, we can use the distance and time data. If the drop's displacement was measured over a 120s interval, the terminal velocity $v_t$ can be calculated as:

$v_t = \frac{d}{t}$

where $d$ is the total displacement measured with the microscope. Using the two positions of the oil drop, we find:

$v_t = \frac{(final position - initial position)}{120}$

Once the specific measurements ('final position' and 'initial position') are identified from the scale provided, we can compute the terminal velocity.

The bar chart displays the charges calculated for fifty oil drops. If the charges predominantly appear as whole multiples of the elementary charge, it would support the student's prediction. However, if the measured values show a distribution that includes fractional or non-integer values, it suggests that other factors are influencing the charge of the oil drops, which may not solely be multiples of the electron charge.