In fair weather, the electric field in the air at a particular location immediately above the Earth's 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 Earth's electric potential due to this charge? (d) What is the difference in potential between the head and the feet of a person 1.75 m tall? (Ignore any charges in the atmosphere.) (e) Imagine the Moon, with 27.3% of the radius of the Earth, had a charge 27.3% as large, with the same sign. Find the electric force the Earth would then exert on the Moon. (f) State how the answer to part (e) compares with the gravitational force the Earth exerts on the Moon.

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.

when the use of electricity burdens the fuse and miniarure circuit beakers and there is a possibility of a fuse then they automatically cut off the electricity supply to your homes and after you fix the problem u jzt switch it on and it continues the supply of electricity to ur homes

a solenoid is created by wrapping a l = 90 m long wire around a hollow tube of diameter d = 4.5 cm.   the wire diameter is d = 0.9 mm.   the solenoid wire is then connected to a power supply so that a current of i = 9 a flows through the wire.

randomized variablesl = 90 m

d = 4.5 cm

d = 0.9 mm

i = 9 a                                          

write an expression for the number of turns, n, in the solenoid. you do not need to take into account the diameter of the wire in this calculation.

calculate the number of turns, n, in the solenoid.

write an expression for the length of the solenoid (l2) in terms of the diameter of the hollow tube d.

calculate the length of the solenoid (l2)   in meters.

calculate the magnitude of the magnetic field at the center of the solenoid in teslas.

explanation: