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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

Principal radii of curvature * Consider a point on the surface of a conductor. The principal radii of curvature of the surface at that point are defined to be the largest and smallest radii of curvature there. To find the radii of curvature, consider a plane that contains the normal to the surface at the given point. Rotate this plane around the normal, and look at the curve representing the intersection of the plane and the surface. The radius of curvature is defined to be the radius of the circle that locally matches up with the curve. For example, a sphere has its principal radii everywhere equal to the radius \(R\). A cylinder has tone principal radius equal to the cross-sectional radius \(R\), and the other equal to infinity. It turns out that the spatial derivative (in the direction of the\\} normal) of the electric field just outside a conductor can be written in terms of the principal radii, \(R_{1}\) and \(R_{2}\), as follows: $$ \frac{d E}{d x}=-\left(\frac{1}{R_{1}}+\frac{1}{R_{2}}\right) E $$ (a) Verify this expression for a sphere, a cylinder, and a plane. (b) Prove this expression. Use Gauss's law with a wisely chosen pillbox just outside the surface. Remember that near the surface, the electric field is normal to it.

Electric field at a corner A very long conducting tube has a square cross section. The charge per unit length in the longitudinal direction is \(\lambda\). Explain why the external electric field diverges at the corners of the tube. Does this result depend on the specific shape of the tube? What if the cross section is triangular or hexagonal? Or what if the point in question is at the tip of a cone or at a kink in a wire?

Capacitance of a spheroid * Here is the exact formula for the capacitance \(C\) of a conductor in the form of a prolate spheroid of length \(2 a\) and diameter \(2 b\) : \(C=\frac{8 \pi \epsilon_{0} a \epsilon}{\ln \left(\frac{1+\epsilon}{1-\epsilon}\right)}, \quad\) where \(\quad \epsilon=\sqrt{1-\frac{b^{2}}{a^{2}}}\) First verify that the formula reduces to the correct expression for the capacitance of a sphere if \(b \rightarrow a\). Now imagine that the spheroid is a charged water drop. If this drop is deformed at constant volume and constant charge \(Q\) from a sphere to a prolate spheroid, will the energy stored in the electric field increase or decrease? (The volume of the spheroid is \((4 / 3) \pi a b^{2}\).)

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?

Infinite square lattice \(*\) * Consider a two-dimensional infinite square lattice of \(1 \Omega\) resistors. That is, every lattice point in the plane has four \(1 \Omega\) resistors connected to it. What is the equivalent resistance between two adjacent nodes? This problem is a startling example of the power of symmetry and superposition. Hint: If you can determine the voltage drop between two adjacent nodes when a current of, say, 1 A goes in one node and comes out the other, then you are done. Consider this setup as the superposition of two other setups.
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