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(a) A general rule for estimating the capacitance \(C\) of an isolated conducting sphere with radius \(r\) is \(C(\) in \(\mathrm{pF}) \approx r(\) in \(\mathrm{cm})\) . That is, the numerical value of \(C\) in \(\mathrm{pF}\) is about the same as the numerical value of the sphere's radius in \(\mathrm{cm} .\) Justify this rule. (b) Modeling the human body as a 1 - 1-m-radius conducting sphere, use the given rule to estimate your body's capacitance. (c) While walking across a carpet, you acquire an excess "static electricity" charge \(Q\) and produce a 0.5 -cm spark when reaching out to touch a metallic door-knob. The dielectric strength of air is 30 \(\mathrm{kV} / \mathrm{cm} .\) Use this information to estimate \(Q(\) in \(\mu \mathrm{C}) .\)

Short Answer

Expert verified
(a) The rule simplifies from physics principles. (b) Capacitance is 100 pF. (c) Charge is approximately 1.5 μC.

Step by step solution

01

Analyze General Capacitance Rule

For a conducting sphere, the capacitance can be calculated using the formula \( C = 4 \pi \varepsilon_0 r \), where \( \varepsilon_0 \) is the permittivity of free space (about \(8.854 \times 10^{-12} \, \text{F/m}\)). In practical units, this simplifies to \( C \approx r \) when \(C\) is expressed in picofarads (pF) and \(r\) in centimeters (cm). This approximation arises because \(4 \pi \varepsilon_0 \approx 1.11 \times 10^{-12} \, \text{F/cm}\), making the capacitance numerically similar to the sphere's radius in cm.
02

Estimate Capacitance of the Human Body

Assuming the human body is a sphere with a radius of 1 meter (or 100 cm), we apply the rule \( C \approx r \). Thus, the capacitance \(C\) is estimated to be approximately 100 pF.
03

Determine the Voltage Across the Spark

The problem states that a spark of 0.5 cm occurs. Given that the dielectric strength of air is 30 kV/cm, the voltage \(V\) associated with this spark is \( V = 30 \, \text{kV/cm} \times 0.5 \, \text{cm} = 15 \, \text{kV} \).
04

Use Capacitance to Find Charge

Using the capacitance equation \( Q = CV \), where \(C = 100 \, \text{pF} = 100 \times 10^{-12} \, \text{F}\) and \( V = 15 \, \text{kV} = 15 \times 10^{3} \, \text{V}\), we calculate the charge: \[Q = 100 \times 10^{-12} \times 15 \times 10^{3} = 1.5 \times 10^{-6} \, \text{C} \\]This charge is equivalent to 1.5 \( \mu \text{C} \).

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Key Concepts

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

Conducting Sphere
When we talk about a conducting sphere in the context of capacitance, we're talking about a perfectly symmetrical object that can store electrical charge. This is because the charge will distribute itself evenly over the surface of the sphere, creating an electric potential. The capacitance of such a sphere is directly related to its size. The larger the sphere, the more charge it can hold for a given electric potential.

The mathematical expression for the capacitance, \( C \), of a conducting sphere can be given by the formula:
  • \( C = 4 \pi \varepsilon_0 r \)
Here, \( \varepsilon_0 \) is the permittivity of free space, and \( r \) is the radius of the sphere. In practical terms, especially in centimeters and picofarads, this boils down to \( C \) being approximately equal to \( r \).

This rule means if the sphere's radius is, say, 10 cm, its capacitance will be approximately 10 pF, making it straightforward to estimate just by knowing the sphere's size.
Dielectric Strength
Dielectric strength is a term used to describe the maximum electric field that a material can withstand without breaking down. In simple terms, it's a measure of a material's insulating strength. For air, this value is around 30 kV/cm. That means you can apply up to 30,000 volts over a one-centimeter air gap before you start to see a spark. This is because the air molecules become ionized, allowing electricity to pass through.

Understanding dielectric strength is essential, especially when working with static electricity and sparks. For example, if you experience a shock while touching a metal object, the spark you see is a result of the air's dielectric breakdown.
Static Electricity
Static electricity refers to the build-up of electrical charge on the surface of objects. This usually happens when two objects are rubbed together, like walking on a carpet. The excess charge stays on the object until it finds a way to discharge, like when you touch a metal doorknob. This discharge takes the form of a spark, as the charge suddenly moves to another object to equalize the potential difference.

In the scenario of walking across a carpet, the human body can become charged due to the friction between the carpet and the shoes. When approaching a conductive object, such as a doorknob, the static charge can cause a small scale lightning bolt, which is otherwise known as a spark.
Permittivity of Free Space
The permittivity of free space, often represented by the symbol \( \varepsilon_0 \), is a fundamental physical constant. It plays a critical role in the equations governing electric fields and capacitance. In simpler terms, it describes how much resistance the vacuum of free space offers against the electric field. Its value is approximately \(8.854 \times 10^{-12} \, \text{F/m}\).

In the context of capacitors, this constant helps us understand how electric fields interact with materials and influences the capacitance value of a sphere or any other object. It is a key factor in the equation for capacitance of a sphere. By knowing this, along with the radius of the sphere, you can accurately calculate the capacitance even in vacuum.

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Most popular questions from this chapter

(II) An electric field of \(4.80 \times 10^{5} \mathrm{V} / \mathrm{m}\) is desired between two parallel plates, each of area 21.0 \(\mathrm{cm}^{2}\) and separated by 0.250 \(\mathrm{cm}\) of air. What charge must be on each plate?

(II) It takes 15 \(\mathrm{J}\) of energy to move a 0.20 \(\mathrm{-mC}\) charge from one plate of a \(15-\mu \mathrm{F}\) capacitor to the other. How much charge is on each plate?

A huge \(3.0-\mathrm{F}\) capacitor has enough stored energy to heat \(3.5 \mathrm{~kg}\) of water from \(22^{\circ} \mathrm{C}\) to \(95^{\circ} \mathrm{C}\). What is the potential difference across the plates?

(II) An electric circuit was accidentally constructed using a \(5.0-\mu \mathrm{F}\) capacitor instead of the required \(16-\mu \mathrm{F}\) value. Without removing the \(5.0-\mu \mathrm{F}\) capacitor, what can a technician add to correct this circuit?

Let us try to estimate the maximum "static electricity" charge that might result during each walking step across an insulating floor. Assume the sole of a person's shoe has area \(A \approx 150 \mathrm{~cm}^{2},\) and when the foot is lifted from the ground during each step, the sole acquires an excess charge \(Q\) from rubbing contact with the floor. (a) Model the sole as a plane conducting surface with \(Q\) uniformly distributed across it as the foot is lifted from the ground. If the dielectric strength of the air between the sole and floor as the foot is lifted is \(E_{\mathrm{S}}=3 \times 10^{6} \mathrm{~N} / \mathrm{C},\) determine \(Q_{\max },\) the maximum possible excess charge that can be transferred to the sole during each step. (b) Modeling a person as an isolated conducting sphere of radius \(r \approx 1 \mathrm{~m},\) estimate a person's capacitance. \((c)\) After lifting the foot from the floor, assume the excess charge \(Q\) quickly redistributes itself over the entire surface area of the person. Estimate the maximum potential difference that the person can develop with respect to the floor.

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