Assume earth and a cloud layer 770 m above the earth can be treated as plates of a parallel-plate capacitor.

a) if the cloud layer has an area of 1 x 106 m2, what is the capacitance?
b) if an electric field strength of 2 x 106 n/c causes the air to conduct charge (lightning), what charge can the cloud hold

(a) 1.15×10⁻⁸ F.

(b) 1.5×10⁻⁸ C.

Step-by-step explanation:

(a)

Capacitance of a capacitor: This is the ability for a capacitor to store charge.

The S.I unit of capacitance of a capacitor is Farad (F).

Mathematically, it is represented as

C = ε₀A/d² Equation 1

Where C = Capacitance of the capacitor, ε₀ = permitivity of free space, A = area of the plates, d = distance between the plates.

Given: A = 1×10⁶ m². d = 770 m

Constant: ε₀ = 8.85×10⁻¹² F/m

Substituting these values into equation 1

C = 8.85×10⁻¹²(1×10⁶ )/770

C = 8.85×10⁻⁶/770

C= 0.0115×10⁻⁶

C = 1.15×10⁻⁸ F.

Thus the capacitance of the cloud layer = 1.15×10⁻⁸ F.

(b)

Electric Field Strength: This is the force per unit charge which is exerted at a point in a in a electric field.

It can be represented mathematically as

E = kq/d² Equation 2

Where E = Electric Field Strength, q = charge, d = distance. k = proportionality constant.

Making q the subject of the equation above,

q = Ed²/k Equation 3.

Given: E = 2×10⁶ N/C, d = 770 m

Constant: k = 1/4πε₀ = 9×10⁹ Nm²/C²

Substituting these values into equation 3

q =  2×10⁶(770)²/(9×10⁹)

q = (2×10⁶)(592900)/(81×10¹⁸)

q = 14639.5×10⁻¹²

q ≈ 1.5×10⁻⁸ C.

Thus the charge the cloud can hold =  1.5×10⁻⁸ C.

ending amount = beginning amount / (2^n)

where "n" = number of half-lives (or elapsed time / half-life)

ending amount = 50 mg / (2 ^10 / 5.3)

ending amount = 50 mg / (2 ^ 1.8867924528)

ending amount = 50 m / 3.6981210597

ending amount = 13.5203794557 mg

or about 13.5 mg remain

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