Answer:
a. 1.062 × 10⁻⁹ C/m² b. 5.472 × 10⁵ C c. 7.685 × 10⁹ V d. -210.12 V e. 4.972 × 10³ N f. F the electric force is much less than the gravitational force F₁ of the earth on the moon.
Explanation:
a. The surface charge density D = ε₀E where E = 120 N/C
D = ε₀E = 8.854 × 10⁻¹² F/m × 120 N/C = 1.062 × 10⁻⁹ C/m²
The charge density is positive since the electric field is positive or directed downwards. Only a positive charge could direct an electric field out of itself.
b. Since the surface charge density is uniform over the planet and charge, Q = DA where A = area of earth = 4πr² where r = radius of earth = 6.4 × 10⁶ m.
So Q = DA
= 1.063 × 10⁻⁹ C/m² × 4π × (6.4 × 10⁶ m)²
= 5.4715 × 10⁵ C
≅ 5.472 × 10⁵ C
c. The earth's electric potential due to this charge Q is V = Q/4πε₀r
= 5.472 × 10⁵ C ÷ (4π × 8.854 × 10⁻¹² F/m × 6.4 × 10⁶ m)
= 5.472 × 10⁵ C ÷ 712.081 × 10⁻⁶ F
= 0.07685 × 10¹¹ V
= 7.685 × 10⁹ V
= 7.685 GV
d. Since V = Q/4πε₀r, and we are going to be dealing with a small change in length compared to r, we differentiate V to get the differential change in potential.
So, dV/dr = -Q/4πε₀r²
dV = -Qdr/4πε₀r² ,dr = 1.75 m
dV = -5.472 × 10⁵ C × 1.75 m ÷ [4π × 8.854 × 10⁻¹² F/m × (6.4 × 10⁶ m)²]
dV = -9.576 × 10⁵ Cm ÷ 4557.318 Fm
dV = -210.124 V
dV ≅ -210.12 V
e. Let Q₁ = charge on moon and Q₂ = charge on earth. Given that Q₁ = 0.273Q₂, the force of attraction between earth and moon is F = Q₁Q₂/4πε₀R² where R = distance between earth and moon = 3.844 × 10⁸ m
F = Q₁Q₂/4πε₀R²
= 0.273Q²/4πε₀R²
= 0.273 × (5.472 × 10⁵ C)²/ [4π × 8.854 × 10⁻¹² F/m × (3.844 × 10⁸ m)²]
= 8.1744 × 10¹⁰ C²/1644.054 Fm × 10⁴
= 0.004972 × 10⁶ N
= 4.972 × 10³ N
= 4.972 kN
f. The gravitational force between earth and moon F₁ = Gm₁m₂/R² where m₁ = mass of moon = 7.3 × 10²² kg and m₂ = mass of earth = 5.97 × 10²⁴ kg.
F₁ = 6.67 × 10⁻¹¹ Nm²/kg² × 5.97 × 10²⁴ kg × 7.3 × 10²² kg/(3.844 × 10⁸ m)²
= 290.685 × 10³⁵ Nm²/14.776 × 10¹⁶ m²
= 19.67 × 10¹⁹ N
= 1.967 × 10²⁰ N
≅ 2 × 10²⁰ N
Since F₁/F = 1.967 × 10²⁰ N/4.972 × 10³ N = 3.95 × 10¹⁶ ≅ 4 × 10¹⁶.
F the electric force is much less than the gravitational force F₁ of the earth on the moon.
