Question

1. Assume that earth has a surface charge density of 1 electron per

metre2

. Calculate the earth’s electric field.

​

Answer 1

ANSWER:

• The earth’s electric field = 1.8 × 10⁻⁸ N/C

GIVEN:

• Earth has a surface charge density of 1 electron per m²

TO FIND:

• The earth’s electric field.

EXPLANATION:

$\blue{ \bigstar \boxed{ \gray{\huge{ \sigma}= \frac{q}{A}}}}$

$\blue{ \bigstar {\boxed{ \gray{\huge{ q= A\sigma}}}}}$

Area of sphere = 4πr²

radius of earth = 6.4 × 10⁶ m

Area of earth = 4 × 3.14 × (6.4 × 10⁶)²

Area of earth = 12.56 × 40.96 × 10¹²

Area of earth = 514.45 × 10¹²

Area of earth = 5.14 × 10¹⁴ m²

σ = - 1.6 × 10⁻¹⁹ C/m²

q = - 1.6 × 10⁻¹⁹ × 5.14 × 10¹⁴

q = - 8.2 × 10⁻⁵ C

$\red{ \bigstar \boxed{ \gray{\huge{E}= \frac{1}{4\pi \epsilon_o}\times \dfrac{q}{r^2}}}}$

$\dfrac{1}{4\pi \epsilon_o} = 9 \times 10^9 \ C^2\ N^{-1} \ m^{-2}$

q = - 8.2 × 10⁻⁵ C

r² = (6.4 × 10⁶)² = 40.96 × 10¹²

$E= \dfrac{9 \times 10^9 \times-8.2\times10^{-5}}{40.96\times10^{12}}$

$E= \dfrac{73.8\times10^{-8}}{40.96}$

E = 1.8 × 10⁻⁸ N/C

Hence the earth’s electric field = 1.8 × 10⁻⁸ N/C