Problem 22 - Fundamentals of Physics

Question

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 \(2.00 \mathrm{~mm}\) extends directly upward by \(0.800 \mathrm{~km}\) and is coated with a \(0.500 \mathrm{~mm}\) layer of water having resistivity \(150 \Omega \cdot \mathrm{m} .\) If the potential difference between the two ends of the string is \(160 \mathrm{MV},\) 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 500000 A (way beyond just being lethal).

Step-by-Step Solution

Answer

The current through the water layer is approximately 1.05 mA.

Step 1: Understand the Problem

We need to find the current flowing through the water coating of the kite string which is subjected to a potential difference. The string acts as a cylinder, and we will use its cylindrical dimensions and electrical properties to solve for the current.

Step 2: Calculate the Cross-Sectional Area

The string is coated with a layer of water, which has a radius of \(2.00 ext{ mm} + 0.500 ext{ mm} = 2.50 ext{ mm} = 0.0025 ext{ m}\). The cross-sectional area \(A\) of the cylindrical water coating is:\[A = \pi (0.0025)^2 - \pi (0.0020)^2\]\[A \approx 7.85 \times 10^{-6} ext{ m}^2\]

Step 3: Calculate the Resistance

Use the resistivity formula for the cylindrical segment:\[R = \frac{\rho L}{A}\]Where:- \(\rho = 150 \, \Omega \cdot \text{m}\) is the resistivity of water,- \(L = 0.800 \, \text{km} = 800 \, \text{m}\) is the length,- \(A\) is the cross-sectional area calculated previously.Substituting the values:\[R = \frac{150 \times 800}{7.85 \times 10^{-6}}\]\[ R \approx 1.53 \times 10^{11} \, \Omega\]

Step 4: Use Ohm's Law to Find the Current

The current \(I\) can be found using Ohm's Law, given by \(V = IR\), where \(V\) is the voltage:\[160 \times 10^6 = I \times 1.53 \times 10^{11}\]Solving for \(I\), we have:\[I = \frac{160 \times 10^6}{1.53 \times 10^{11}}\]\[I \approx 1.05 \times 10^{-3} \, \text{A}\]

Step 5: Conclusion and Summary

The calculated current is approximately \(1.05 \times 10^{-3} \, \text{A}\), or \(1.05 \text{ mA}\). This is a small current compared to the potential lightning strike hazard deemed dangerous at 500000 A.

Key Concepts

Ohm's Law

Ohm's Law is a fundamental principle in the study of electricity that relates voltage, current, and resistance in an electrical circuit. It states that the current (\(I\)) flowing through a conductor between two points is directly proportional to the voltage (\(V\)) across the two points, and inversely proportional to the resistance (\(R\)) of the conductor. This relationship can be represented by the formula:

• \( V = IR \)

This means if you know any two of these variables, you can calculate the third. In the context of the problem, Ohm's Law was employed to find the current through the water layer on the kite string when a potential difference of 160 MV is applied. This foundational concept guides engineers and electricians in designing and troubleshooting electrical circuits by helping determine how much current will flow in a part of a circuit based on the known voltage and resistance. Understanding Ohm's Law is crucial because it also allows predicting how changes in voltage or resistance impact the current.

Resistivity

Resistivity is a property of materials that measures how strongly a material resists the flow of electric current. It is denoted by the Greek letter rho (\( \rho \)) and is typically expressed in ohm-meters (\( \Omega \cdot m \)). The lower the resistivity, the more easily an electric current can flow through the material.
The resistivity of a material depends not only on the material itself but also on environmental conditions such as temperature and humidity. In the example with the kite string, water was the material coating the string, and it had a resistivity of 150 \( \Omega \cdot m \). Matching the resistivity of a material with the right application is key in engineering decisions for effective electrical designs.
For example:- Conductive materials with low resistivity like copper are used in wires.- Insulating materials with high resistivity are used to prevent unwanted flow of current.Within the problem's context, resistivity was integral in calculating the resistance of the water layer using the formula \( R = \frac{\rho L}{A} \), where \( L \) is the length and \( A \) is the cross-sectional area.

Cylindrical Conductors

When dealing with cylindrical conductors, their geometry greatly impacts how electrical calculations are performed. A cylindrical conductor is essentially a symmetrical shape with a specific radius and length. Calculating properties like resistance or current in such a conductor requires understanding its volume and surface area.
For example, in this exercise, the kite string is treated as a cylindrical conductor coated with water. To find the resistance, we first needed the cross-sectional area of the water layer, calculated through the formula for the area of a circle: \(A = \pi r^2\).Here, the conductance calculation involved the radii of the kite string and the water coating.
The steps include:

• Determining the total radius (\(2.50 \) mm) and the core radius (\(2.00 \) mm).
• Using these to find the area of the water layer, which dictates the path available for current flow.

Such considerations ensure that electrical properties are calculated accurately, revealing how crucial understanding shape and dimensions are in the practical application of electrical principles.

Potential Difference

Potential difference, often referred to as voltage, is a measure of the electric potential energy between two points in a circuit. It is essentially the driving force that pushes or pulls electric charges through a conductor and is measured in volts (\( V \)).
In the context of the given problem, a potential difference of 160 million volts is applied across the kite string. This enormous voltage is what creates the electric current to flow through the water layer on the string despite its resistance.
Understanding potential difference is critical because it determines how much work can be done by electric charges. In practical terms, it is what powers electronic devices and systems.

• A high potential difference can drive a lot of current through a low resistance, resulting in more power and sometimes danger, as seen with the risk of lightning strikes.
• Conversely, a low potential difference results in less current and power.

Recognizing and controlling potential difference helps in designing circuits to function safely and effectively, preventing overloads or failures.