Physics 9702/12 — May/June 2019

Background Concept

SI prefixes represent powers of ten:

• $\mu$ (micro) means $10^{-6}$
• $\text{m}$ (milli) means $10^{-3}$

So converting from micro to milli changes the power of ten by a factor of $10^3$.

Also, potential difference is defined as energy transferred per unit charge:

$$V = \frac{W}{Q}$$

So $1\,\text{V} \equiv 1\,\text{J C}^{-1}$.

Understanding the Question

You are given a voltage written with the prefix micro: $2000\,\mu\text{V}$. The task is to rewrite it in an equivalent form and choose which option matches exactly.

The options include different prefixes (milli, pico) and one option written in base units $\text{J C}^{-1}$.

Approach

1. Convert $2000\,\mu\text{V}$ into volts using $\mu = 10^{-6}$.
2. Rewrite the result using a convenient prefix (likely millivolts).
3. Match the numerical value and unit to the options.

Step-by-Step Reasoning

Start with the prefix conversion:

$$2000\,\mu\text{V} = 2000 \times 10^{-6}\,\text{V}$$

Calculate the power of ten:

$$2000 \times 10^{-6} = 2.0 \times 10^{3} \times 10^{-6} = 2.0 \times 10^{-3}$$

So:

$$2000\,\mu\text{V} = 2.0 \times 10^{-3}\,\text{V}$$

Now use $1\,\text{mV} = 10^{-3}\,\text{V}$:

$$2.0 \times 10^{-3}\,\text{V} = 2.0\,\text{mV}$$

This matches option B.

(Checking the tempting unit option: since $\text{V} = \text{J C}^{-1}$, then $2\,\mu\text{J C}^{-1} = 2\,\mu\text{V}$, not $2000\,\mu\text{V}$.)

Key Takeaways

• Know common SI prefixes: $\mu = 10^{-6}$, $\text{m} = 10^{-3}$, $\text{p} = 10^{-12}$.
• Convert by writing the prefix as a power of ten, then simplify.
• Remember $\text{V} \equiv \text{J C}^{-1}$.

Common Mistakes

• Treating $\mu$ as $10^{-3}$ (confusing micro with milli).
• Moving the decimal the wrong way when converting to millivolts.
• Thinking that any $\text{J C}^{-1}$ value is automatically correct without matching the numerical factor.

Things to Be Careful About

• $2000\,\mu\text{V}$ is larger than $2\,\mu\text{V}$ by a factor of $1000$.
• When rewriting in millivolts, the number should become smaller than in microvolts because $\text{mV}$ is a larger unit than $\mu\text{V}$.
• Ensure the final matches both the unit and the magnitude (including the prefix).