Question

p55. in fair weather, the electric field in the air at a particular location immediately above the earths surface is 120 n/c directed downward, (a) what is the surface charge density on the ground? is it positive or negative? (b) imagine the surface charge density is uniform over the planet. what then is the charge of the whole surface of the earth? (c) what is the earths electric potential due to this charge?

Explanation:

Identify given values and constants

We are given:

• Electric field at the surface: \(E = 120\text{ N/C}\) directed downward (toward the center of the Earth).
• Permittivity of free space: \(\varepsilon_0 \approx 8.854 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2)\).
• Radius of the Earth: \(R \approx 6.37 \times 10^6\text{ m}\).
• Coulomb's constant: \(k_e = \frac{1}{4\pi\varepsilon_0} \approx 8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2\).

Determine surface charge density

Using the Electrostatics knowledge point

$$ E = \frac{|\sigma|}{\varepsilon_0} \implies |\sigma| = \varepsilon_0 E $$

$$ |\sigma| = (8.854 \times 10^{-12}\text{ C}^2/(\text{N}\cdot\text{m}^2))(120\text{ N/C}) \approx 1.06 \times 10^{-9}\text{ C/m}^2 $$

Since the electric field points downward (toward the ground), the charge on the ground must be negative:

$$ \sigma \approx -1.06 \times 10^{-9}\text{ C/m}^2 $$

Calculate total charge of the Earth

Using the Electrostatics knowledge point

$$ Q = \sigma A = \sigma (4\pi R^2) $$

$$ Q = (-1.062 \times 10^{-9}\text{ C/m}^2) \cdot 4\pi (6.37 \times 10^6\text{ m})^2 \approx -5.42 \times 10^5\text{ C} $$

Calculate Earth's electric potential

Using the Electric Potential knowledge point

$$ V = \frac{k_e Q}{R} $$

$$ V = \frac{(8.99 \times 10^9\text{ N}\cdot\text{m}^2/\text{C}^2)(-5.42 \times 10^5\text{ C})}{6.37 \times 10^6\text{ m}} \approx -7.65 \times 10^8\text{ V} $$

Answer:

(a) The surface charge density on the ground is \(-1.06 \times 10^{-9}\text{ C/m}^2\). It is negative.

(b) The total charge of the whole surface of the Earth is \(-5.42 \times 10^5\text{ C}\).

(c) The Earth's electric potential due to this charge is \(-7.65 \times 10^8\text{ V}\).