2020 A/L Physics Past Paper · Question 31 (English Medium)

Answer: (4) $100\ \text{V}$

What the question asks

We must find: the potential $V'$ of one drop formed by merging $10^6$ droplets; each droplet's potential is $V = 0.01$ V.

Relevant theory

When a conducting sphere is given an electric charge, we already know that this will reside only on the outer surface of the sphere.

$$V = \left(\frac{1}{4\pi\varepsilon}\right)\frac{Q}{r}$$

Required equations

$V = \frac{1}{4\pi\varepsilon_0}\frac{q}{r} = 0.01\ \text{V}$

$\frac{4}{3}\pi R^3 = 10^6 \times \frac{4}{3}\pi r^3$

$Q = 10^6\,q$

$V' = \frac{1}{4\pi\varepsilon_0}\frac{Q}{R}$

Reasoning and calculation

What changes and what stays the same: the total volume of mercury and the total charge do not change — nothing is lost as the droplets merge.

What changes: the radius $r \to R$, and the charge on one sphere $q \to Q = 10^6 q$. The constant $\dfrac{1}{4\pi\varepsilon_0}$ is common to both states.

$R^3 = 10^6\,r^3$

$100$

$R = \sqrt[3]{10^6}\,r = 100\,r$

$\frac{V'}{V} = \frac{Q/R}{q/r} = \frac{Q}{q} \times \frac{r}{R}$

$\frac{V'}{V} = 10^6 \times \frac{1}{100}$

$10^4$

$\frac{V'}{V} = 10^4$

$V' = 10^4 \times 0.01$

Reading the options

Comparing with the options: $V' = 100$ V — option (4).

Checks

$V0 * pow(N, 2 / 3)$

Dimension check: $\dfrac{V'}{V} = \dfrac{Q}{q}\cdot\dfrac{r}{R}$ is dimensionless; so $V'$ has the same unit as $V$ (volt) ✓.

Second way (general formula): for $N$ droplets, $V' = N^{2/3}V$; if $N = 1$ then $V' = V$ ✓ (a boundary check), and if $N = 10^6$ then $(10^6)^{2/3} = 10^4$ ✓.

Conclusion

$V' = 100\ \text{V}$

The charge grows $10^6$ times, the radius $100$ times: $V' = \dfrac{10^6}{100} \times 0.01 = 100\ \text{V}$ — the answer is (4).

Taking a square root, or the wrong cube root, to find the radius.

Volume $\propto R^3$: $R = \sqrt[3]{10^6}\,r = 100\,r$.