You are working as an intern for a meteorological laboratory. You are out in the field taking measurements from a balloon that is carrying equipment designed to measure electric fields in the atmosphere. Your supervisor has asked you to determine the average volume charge density at a certain height in the air. When the balloon is at an altitude of 500 m above the ground, the electric field is measured to be 160 N/C directed downward. At 800 m above the ground, the electric field is 120 N/C downward. (a) Determine the average volume charge density (in C/m3) in the layer of air between these two elevations. (Enter the magnitude.)
(b) Is this layer of air positively or negatively charged?
positively charged
negatively charged

a.  -6.99 × 10³⁰ C/m³ b. The layer of air is negatively charged.

Explanation:

With E₁ = electric field at 500 m above the ground = 160 N/C (it is negative since it is directed downwards).

Also, with E₂ = electric field at 800 m above the ground = 120 N/C (it is negative since it is directed downwards).

The total flux, Ψ = ∫E.dA = E₁dAcosθ + E₂dAcosθ

For the 800 m surface E is parallel to dA, that is = 0° and For the 500 m surface E is anti-parallel to dA, that is = 180°  

Ψ = ∫E₁dAcos180° + ∫E₂dAcos0°

= -∫E₁dA + ∫E₂dA

= -E₁∫dA + E₂∫dA

= -E₁4πR² + E₂4πR²

= (-E₁ + E₂)4πR² where R = radius of earth = 6.4 × 10⁶ m

= (-160 N/C + 120 N/C)4π(6.4 × 10⁶ m)²

= - 40 N/C)4π(6.4 × 10⁶ m)²

= -20588.74 × 10¹² C

= -2.058874 × 10¹⁶ C

≅ -2.06 × 10¹⁶ Nm²/C.

The since charge, Q = Ψ/ε₀, the total charge through the area is thus

Q = Ψ/ε₀

= -2.06 × 10¹⁶  Nm²/C ÷ 8.854 × 10⁻¹² F/m

= -0.233 × 10²⁸ C/m²

= -2.33 × 10²⁸ C/m².

So, the charge in the volume = charge net charge of surface × width of volume. So the charge in the volume Q' = QΔh = Q(h₂ - h₁) where h₁ = 500 m and h₂ = 800 m

Q' = Q(h₂ - h₁)

=  -2.33 × 10²⁸ C/m²(800 m - 500 m)

= -2.33 × 10²⁸ C/m²(300 m)

= -699 × 10²⁸ C/m³

= -6.99 × 10³⁰ C/m³

b. Since Q' = -6.99 × 10³⁰ C/m³, the layer of air is negatively charged.

b would be the answer