Kiting during a storm. The legend that Benjamin Franklin flew a kite as a storm approached is only a legend-he was neither stupid nor suicidal. Suppose a kite string of radius 2.00 mm extends directly upward by 1.80 km and is coated with a 0.500 mm layer of water having resistivity `150Omega.m.` If the potential difference between the two ends of the string is 213 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).

Answer

To solve the problem, we need to calculate the current through the water layer coating the kite string using the given values. We will follow these steps:

### Step 1: Identify the Given Values
- Radius of the kite string, \( r = 2.00 \, \text{mm} = 2.00 \times 10^{-3} \, \text{m} \)
- Thickness of the water layer, \( t = 0.500 \, \text{mm} = 0.500 \times 10^{-3} \, \text{m} \)
- Length of the string, \( L = 1.80 \, \text{km} = 1800 \, \text{m} \)
- Resistivity of water, \( \rho = 150 \, \Omega \cdot \text{m} \)
- Potential difference, \( V = 213 \, \text{MV} = 213 \times 10^{6} \, \text{V} \)

### Step 2: Calculate the Cross-Sectional Area of the Water Layer
The cross-sectional area \( A \) of the water layer can be calculated using the formula:
\[ A = \pi \left( (r + t)^2 - r^2 \right) \]
Substituting the values:
\[ A = \pi \left( (2.00 \times 10^{-3} + 0.500 \times 10^{-3})^2 - (2.00 \times 10^{-3})^2 \right) \]
Calculating \( r + t \):
\[ r + t = 2.00 \times 10^{-3} + 0.500 \times 10^{-3} = 2.50 \times 10^{-3} \, \text{m} \]
Now calculating \( A \):
\[ A = \pi \left( (2.50 \times 10^{-3})^2 - (2.00 \times 10^{-3})^2 \right) \]
\[ = \pi \left( 6.25 \times 10^{-6} - 4.00 \times 10^{-6} \right) \]
\[ = \pi \times 2.25 \times 10^{-6} \approx 7.07 \times 10^{-6} \, \text{m}^2 \]

### Step 3: Calculate the Resistance of the Water Layer
Using the formula for resistance:
\[ R = \frac{\rho L}{A} \]
Substituting the values:
\[ R = \frac{150 \, \Omega \cdot \text{m} \times 1800 \, \text{m}}{7.07 \times 10^{-6} \, \text{m}^2} \]
Calculating \( R \):
\[ R \approx \frac{270000 \, \Omega \cdot \text{m}}{7.07 \times 10^{-6}} \approx 3.81 \times 10^{10} \, \Omega \]

### Step 4: Calculate the Current through the Water Layer
Using Ohm's law:
\[ I = \frac{V}{R} \]
Substituting the values:
\[ I = \frac{213 \times 10^{6} \, \text{V}}{3.81 \times 10^{10} \, \Omega} \]
Calculating \( I \):
\[ I \approx 5.58 \times 10^{-3} \, \text{A} \]

### Final Answer
The current through the water layer is approximately:
\[ I \approx 5.58 \, \text{mA} \]

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