91Ó°ÊÓ

In places such as a hospital operating room and a factory for electronic circuit boards, electric sparks must be avoided. A person standing on a grounded floor and touching nothing else can typically have a body capacitance of \(150 \mathrm{pF}\), in parallel with a foot capacitance of \(80.0 \mathrm{pF}\) produced by the dielectric soles of his or her shoes. The person acquires static electric charge from interactions with furniture, clothing, equipment, packaging materials, and essentially everything else. The static charge is conducted to ground through the equivalent resistance of the two shoe soles in parallel with each other. A pair of rubber-soled street shoes can present an equivalent resistance of 5000 M\Omega. A pair of shoes with special static-dissipative soles can have an equivalent resistance of 1.00 M\Omega. Consider the person's body and shoes as forming an \(R C\) circuit with the ground. (a) How long does it take the rubber-soled shoes to reduce a \(3000-\mathrm{V}\) static charge to \(100 \mathrm{V} ?\) (b) How long does it take the staticdissipative shoes to do the same thing?

Step by step solution

01

- Calculate the total capacitance

The given values are 150pF and 80.0pF which when combined in parallel, adds up. Hence, \(C = C1+C2 = 150pF+80pF = 230pF\) or \(230 x 10^{-12} F\).

02

- Calculate time constant \(\tau\) for rubber-soled shoes

The time constant \(\tau = RC\) is used to calculate how fast the system charges or discharges. The resistance for rubber-soled shoes is \(R = 5000M\Omega\) or \(5000 x 10^6 \Omega\). Therefore, \(\tau = RC = (5000 x 10^6 \Omega)(230 x 10^{-12} F) = 1.15s\).

03

- Calculate the discharge time for rubber-soled shoes

The formula used to calculate the time it takes to discharge a capacitor to a certain voltage is \[t = -\tau \ln(V/V_0)\] where V is the final voltage, V0 is the initial voltage, and \(\tau\) is the time constant. The time it takes for the static charge to reduce from 3000V to 100V is then calculated as \(t = -\tau \ln(V/V_0) = -1.15s \ln(100V/3000V) = 4.48s\).

04

- Calculate time constant \(\tau\) for static-dissipative shoes

The resistance for static-dissipative shoes is \(R = 1M\Omega\) or \(1 x 10^6 \Omega\). Therefore, time constant \(\tau = RC = (1 x 10^6 \Omega)(230 x 10^{-12} F) = 0.23s\).

05

- Calculate the discharge time for static-dissipative shoes

Using the same discharge formula, the time it takes for the static charge to reduce from 3000V to 100V is calculated as \(t = -\tau \ln(V/V_0) = -0.23s \ln(100V/3000V) = 0.897s\) .

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding the Time Constant in an RC Circuit

In an RC circuit, knowing the time constant is crucial for understanding how quickly the circuit can charge or discharge. The time constant, represented by the Greek letter \( \tau \), is the product of the resistance \( R \) and capacitance \( C \) of the circuit: \( \tau = RC \). This value tells us the time it takes for the charge in the circuit to drop to about 37% of its initial value in the process of discharge.

In practical terms:

• The larger the time constant, the slower the discharge of the capacitor will be.
• A small time constant means a faster discharge.

For our problem, both rubber-soled shoes and static-dissipative shoes create different time constants due to their different resistances. Calculating \( \tau \) allows us to evaluate how the type of shoes affects the discharge speed from a static shock. The rubber-soled shoes have a larger \( \tau \), leading to a slower discharge compared to the static-dissipative shoes with their smaller \( \tau \). This illustrates how varying components in an RC circuit can significantly alter its behavior.

The Role of Body Capacitance

In electrotechnics, body capacitance is a critical concept, particularly in environments where static electricity needs to be managed carefully, like in hospitals and electronic factories. Capacitance is the ability of a system to store an electric charge. When a person is on a grounded surface, their capacitance can be likened to a component in an electronic circuit.

In this exercise, we combined:

• A body capacitance of \(150 \mathrm{pF}\)
• A foot capacitance produced by shoe soles of \(80.0 \mathrm{pF}\)

As they are in parallel, they sum up to a total capacitance of \(230 \mathrm{pF}\). What does this mean? In our context, the higher body capacitance increases the overall capacity for storing charge, making discharge times longer. This understanding is key when analyzing how quickly a person can become neutralized from static charge, characteristic in circuit design where quick discharge is often necessary.

Body capacitance, in conjunction with the resistance offered by shoes, forms the complete picture of how static electricity dissipates in a person acting as an RC circuit.

Calculating Discharge Time

The discharge time in an RC circuit is the interval required for the voltage across the capacitor to decrease to a specific level. This is modeled in our exercise by the equation:\[ t = -\tau \ln\left(\frac{V}{V_0}\right) \]This formula can be used to find how long it takes for the voltage to fall from an initial value \( V_0 \) to a final value \( V \).

For instance, we calculated the time to reduce a 3000V static charge to 100V. By inputting the time constant \( \tau \) into the equation for both the rubber-soled and static-dissipative shoes:

• Rubber-soled shoes: \( \tau = 1.15s \) leading to a discharge time of 4.48 seconds.
• Static-dissipative shoes: \( \tau = 0.23s \) leading to a much faster discharge time of 0.897 seconds.

This demonstrates the efficiency of different materials and designs in reducing static charge, essential in environments where electrical safety is paramount. Discharge time is an essential aspect of how components in an RC circuit interact to meet specific needs.