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Chapter 3: Problem 1

Van de Graaff current ? In a Van de Graaff electrostatic generator, a rubberized belt \(0.3 \mathrm{~m}\) wide travels at a velocity of \(20 \mathrm{~m} / \mathrm{s}\). The belt is given a surface charge at the lower roller, the surface charge density being high enough to cause a field of \(10^{6} \mathrm{~V} / \mathrm{m}\) on each side of the belt. What is the current in milliamps?

Short Answer

Expert verified
The current produced by the Van De Graaff generator would be approximately 0.00354 milliamperes.

Step by step solution

01

Identify Relevant Data

The electric field \(E\) is given as \(10^6 V/m\). The belt's width \(w\) is \(0.3 m\) and its velocity \(v\) is \(20 m/s\). Most importantly, the problem provides that the field is on each side of the belt, meaning we have two sides carrying charge.
02

Calculate Charge

Knowing that the electric field \(E\) is given as \(10^6 V/m\), and we also know that \(E=\sigma / \varepsilon_0\) (where \(\sigma\) is surface charge density and \(\varepsilon_0\) is the permittivity of free space which is \(8.85 脳 10^{-12} C^2/N \cdot m^2\)), so we can use this to calculate the surface charge density \(\sigma\). Thus, we have \(\sigma=E \times \varepsilon_0 = 10^6 \times 8.85 脳 10^{-12} C/m^2 = 8.85 \times 10^{-6} C/m^2 \). Since the field is on both sides of the belt, the total charge per unit area is \(2\sigma = 2*8.85 脳 10^{-6} C/m^2 = 1.77 脳 10^{-5} C/m^2\).
03

Calculate Current

The rate of flow of these charges or the current is calculated as \(I = nVdq/dt\), with \(n = 2\sigma\) (from previous calculation), \(V = 20 m/s\) and \(dq/dt = nw\), where \(w\) is the width of the belt. Substituting these values, we get \( I = nV dq/dt = 2\sigma * v * w = 2 * (1.77 脳 10^{-5} C/m^2) * (20 m/s) * (0.3 m) = 2 * (1.77 脳 10^{-6} A) = 3.54 脳 10^{-6} A \). Since 1 Ampere = 1000 milliamperes, the current value in milliamperes is \(3.54 脳 10^{-6} A * 1000 = 0.00354 mA\).

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Key Concepts

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

Surface Charge Density
Surface charge density is a term that describes the amount of electric charge per unit area on a surface. In the case of a Van de Graaff generator, the rubberized belt carries a certain charge as it moves. This charge is distributed over the area of the belt, creating a surface charge density. - Surface charge density is commonly denoted by the symbol \(\sigma\).- It is measured in coulombs per square meter (C/m虏).Given the electric field \(E\) produced on the belt is \(10^6 \, \mathrm{V/m}\), and knowing the relationship \(E = \sigma / \varepsilon_0\), the surface charge density \(\sigma\) can be calculated by rearranging this formula to \(\sigma = E \cdot \varepsilon_0\). Here, \(\varepsilon_0\) is the permittivity of free space, a constant essential in electrostatics. This relationship highlights how the electric field on the belt relates directly to the amount of charge it carries.
Electric Field
The electric field is a vector field around a charged object where forces are exerted on other charges. In this exercise, the electric field is given as \(10^6 \, \mathrm{V/m}\). This field is substantial and contributes to the behavior of the Van de Graaff generator.- The electric field \(E\) is measured in volts per meter (V/m).- It shows how the charge affects the space around it.When considering two sides of a belt carrying charge, the electric field influences how we calculate the total charge and, consequently, the current. It serves as a bridge between the surface charge density and the resulting current, allowing us to understand how charges move and interact through the machine's operation.
Permittivity of Free Space
Permittivity of free space, represented by \(\varepsilon_0\), is a fundamental physical constant important in electrostatics. It quantifies the ability of a vacuum to permit electric field lines. This constant plays a crucial role in calculations involving electric fields and charge distributions.- The standard value is \(\varepsilon_0 = 8.85 \times 10^{-12} \, \mathrm{C^2/N \, \cdot \, m^2}\).- It is essential for determining relationships between electric fields and the charge density \(\sigma\).In the context of the Van de Graaff generator, \(\varepsilon_0\) helps calculate the surface charge density, linking the known electric field to the charge on the belt through the expression \(\sigma = E \cdot \varepsilon_0\). Understanding \(\varepsilon_0\) aids in comprehending how electric fields behave in space free of any physical material.
Current Calculation
Current calculation involves determining how much electric charge flows through a conductor over a time period. In the case of the Van de Graaff generator, it means calculating the current produced by the movement of charge on the belt.Current \(I\) is derived by looking at the rate of charge flow. The problem uses parameters such as:- Belt velocity \(v = 20 \, \mathrm{m/s}\).- Belt width \(w = 0.3 \, \mathrm{m}\).- Total charge per unit area \(2\sigma\).Utilizing the formula \(I = 2\sigma \times v \times w\), you can determine the current in amperes. Here, \(2\sigma\) reflects the charge on both sides of the belt. Translating this result into milliamperes involves converting from amperes by multiplying by 1000, helping students understand the flow of charges in an accessible manner.

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Most popular questions from this chapter

Capacitance of raindrops \(N\) charged raindrops with radius \(a\) all have the same potential. Assume that they are far enough apart so that the charge distribution on each isn't affected by the others (that is, it is spherically symmetric). What is the total capacitance of this system? How does this capacitance compare with the capacitance in the case where the drops are combined into one big drop?

Two capacitors with the same capacitance \(C\) and charge \(Q\) are placed next to each other, as shown in Fig. 3.39. The two positive plates are then connected by a wire. Will charge flow in the wire? Consider two possible reasonings: (A) Before the plates are connected, the potential differences of the two capacitors are the same (because \(Q\) and \(C\) are the same). So the potentials of the two positive plates are equal. Therefore, no charge will flow in the wire when the plates are connected. (B) Number the plates 1 through 4 , from left to right. Before the plates are connected, there is zero electric field in the region between the capacitors, so plate 3 must be at the same potential as plate 2 . But plate 2 is at a lower potential than plate \(1 .\) Therefore, plate 3 is at a lower potential than plate 1 , so charge will flow in the wire when the plates are connected. Which reasoning is correct, and what is wrong with the wrong reasoning?

Distribution of charge on a capacitor Consider a parallel-plate capacitor with different magnitudes of charge on the two plates. Let the charges be \(Q_{1}\) and \(Q_{2}\) (which we normally set equal to \(Q\) and \(-Q\) ). Find the four amounts of charge on the inner and outer surfaces of the two plates.

Conductor in a capacitor (a) The plates of a capacitor have area \(A\) and separation \(s\) (assumed to be small). The plates are isolated, so the charges on them remain constant; the charge densities are \(\pm \sigma .\) A neutral conducting slab with the same area \(A\) but thickness \(s / 2\) is initially held outside the capacitor, see Fig. \(3.40 .\) The slab is released. What is its kinetic energy at the moment it is completely inside the capacitor? (The slab will indeed get drawn into the capacitor, as evidenced by the fact that the kinetic energy you calculate will be positive.) (b) Same question, but now let the plates be connected to a battery that maintains a constant potential difference. The charge densities are initially \(\pm \sigma\). (Don't forget to include the work done by the battery, which you will find to be nonzero.)

Image charges for two planes : A point charge \(q\) is located between two parallel infinite conducting planes, a distance \(d\) from one and \(\ell-d\) from the other. Where should image charges be located so that the electric field is everywhere perpendicular to the planes?
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