Question

Assume that the surface charge density of the Earth is 1 electron charge/m². The potential of the Earth's surface (radius 6400 km) will be:

• A. $-0.12\,\text{V}$
• B. $-0.21\,\text{V}$
• C. $+2.1\,\text{V}$
• D. $+1.2\,\text{V}$

Hint:

Use \(V=\sigma R/\varepsilon_0\) with \(\sigma=-1.6\times10^{-19}\) C/m\(^2\) and \(R=6.4\times10^{6}\) m. The electron charge makes it negative.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
The Earth is treated as a uniformly charged conducting sphere. The charge on it is spread evenly over the surface with surface charge density \(\sigma\). An electron has a negative charge, so the Earth carries a negative charge here.

Step 2: Key Formula or Approach:
For a charged sphere of radius \(R\), the potential on its surface is \(V=\dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{R}\). The total charge is \(Q=\sigma\cdot4\pi R^{2}\).
Put the two together: \[ V=\frac{\sigma\,4\pi R^{2}}{4\pi\varepsilon_0R}=\frac{\sigma R}{\varepsilon_0} \]

Step 3: Detailed Explanation:
Here \(\sigma=-1.6\times10^{-19}\ \text{C m}^{-2}\), \(R=6400\ \text{km}=6.4\times10^{6}\ \text{m}\) and \(\varepsilon_0=8.85\times10^{-12}\ \text{C}^{2}\text{N}^{-1}\text{m}^{-2}\).
\[ V=\frac{(-1.6\times10^{-19})(6.4\times10^{6})}{8.85\times10^{-12}}=\frac{-1.024\times10^{-12}}{8.85\times10^{-12}}\approx-0.12\ \text{V} \]

Step 4: Why the other options are wrong:
Options (C), \(+2.1\) V, and (D), \(+1.2\) V, are positive. A surface charged with electrons must give a negative potential, so both are wrong. Option (B), \(-0.21\) V, has the right sign but is almost double the calculated value, so it is wrong too.

Final Answer:
The potential of the Earth's surface is about \(-0.12\) V, which is option (A). \[ \boxed{V\approx-0.12\ \text{V}} \]