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An irregular neutral conductor has a hollow cavity inside of it and is insulated from its surroundings. An excess charge of \(+16 \mathrm{nC}\) is sprayed onto this conductor. (a) Find the charge on the inner and outer surfaces of the conductor. (b) Without touching the conductor, a charge of \(-11 \mathrm{nC}\) is inserted into the cavity through a small hole in the conductor. Find the charge on the inner and outer surfaces of the conductor in this case.

Short Answer

Expert verified
Initial: Inner = 0 nC, Outer = +16 nC; After insertion: Inner = +11 nC, Outer = +5 nC.

Step by step solution

01

Understanding Charge Distribution on the Conductor

In the initial case, before any charge is inserted into the cavity, the entire excess charge of \(+16\, \text{nC}\) is distributed on the outer surface of the conductor. The inner surface remains with zero net charge because no charge is inside the cavity to induce any charge on the inner surface.
02

Effect of External Charge Introduction in the Cavity

When a charge of \(-11\, \text{nC}\) is introduced into the cavity, according to Gauss's Law, the inner surface must have a charge equal and opposite to that of the charge inside the cavity, i.e., \(+11\, \text{nC}\). This ensures that the electric field inside the conductor material itself remains zero.
03

Calculating the Remaining Charge on the Outer Surface

The total charge originally on the conductor was \(+16\, \text{nC}\). With the inner surface now having \(+11\, \text{nC}\), the outer surface has to account for the rest of the charge to maintain the total at \(+16\, \text{nC}\). Therefore, the outer surface will have \(+16\, \text{nC} - (+11\, \text{nC}) = +5\, \text{nC}\).

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Key Concepts

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

Gauss's Law
Gauss's Law is a fundamental principle of electrostatics. It relates the distribution of electric charge to the resulting electric field. The law states that the net electric flux through a closed surface is directly proportional to the enclosed electric charge. In mathematical form, Gauss's Law is expressed as: \[ \Phi = \oint \mathbf{E} \cdot \text{d}\mathbf{A} = \frac{Q_{\text{enc}}}{\varepsilon_0}, \] where \( \Phi \) is the electric flux, \( \mathbf{E} \) is the electric field, \( \text{d}\mathbf{A} \) is the differential area on the closed surface, \( Q_{\text{enc}} \) is the enclosed charge, and \( \varepsilon_0 \) is the permittivity of free space.
Gauss's Law helps us understand the behavior of electric fields within and around conductors. For a conductor in electrostatic equilibrium, the electric field inside the material is zero. When we introduce a charge within a cavity of a conductor, Gauss's Law allows us to determine the induced charge on the cavity's surface by setting the net electric flux inside the conductor to zero.
Charge Distribution
Charge distribution refers to the way excess electric charges are spread out over a conductor. In a conductor, free charges move until they reach a state where there is no net force acting on them. Thus, these charges reside only on the surface of conductors in electrostatic equilibrium.
The distribution of charge on the surface depends on the shape of the conductor. On an irregularly shaped conductor, charges tend to accumulate at points of high curvature or sharp edges. This minimizes repulsion between the charges as they attempt to spread out as evenly as possible due to mutual repulsion.
  • For a charged conductor, excess charge resides solely on its outer surface when it is isolated from other charges.
  • If a cavity within the conductor contains a charge, the conductor's inner surface will distribute a charge equal and opposite to ensure the electric field inside the conductor material itself remains zero.
Understanding charge distribution is crucial in solving problems involving conductors, especially when dealing with excess charge or external influences such as other nearby charges.
Conductors and Insulators
Conductors and insulators are materials with distinct electrical properties. Conductors, such as metals, allow electric charge to flow freely because they contain a high density of charge carriers, usually electrons. Conversely, insulators, like rubber or glass, have very few charge carriers and do not conduct electricity well.
In electrostatics, conductors are especially important because they have properties that dictate charge distribution. Some of these key properties include:
  • Equal Potential: The entire conductor is at the same electric potential, meaning there are no electric fields within the conductor in electrostatic equilibrium.
  • Surface Charges: Conductor surfaces can hold excess charge because electromagnetic interactions force charges to distribute themselves along the surface.
An understanding of these properties is crucial for grasping how conductors respond to external fields and charges. Insulators, on the other hand, maintain charges in static positions because electricity does not easily flow through them.
Electric Field
The electric field is a vector field around charged objects where forces are exerted on other charges. It provides a way to visualize how electric forces act over a distance. The electric field \( \mathbf{E} \) is defined as the force \( \mathbf{F} \) per unit charge \( q \): \[ \mathbf{E} = \frac{\mathbf{F}}{q}. \]
In electrostatics, the electric field is critical for understanding interactions between charges and how they influence the environment around them. The electric field's magnitude and direction can be calculated using Gauss’s Law and other principles:
  • Inside a conductor in equilibrium, the electric field is zero. This is due to the redistribution of charges, which cancels internal electric fields.
  • Just outside a charged conductor, the electric field is perpendicular to the surface and proportional to the surface charge density.
Recognizing these key characteristics of electric fields helps in solving problems related to charge distribution and electrostatic equilibrium.

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Most popular questions from this chapter

(a) A closed surface encloses a net charge of \(2.50 \mu \mathrm{C}\). What is the net electric flux through the surface? (b) If the electric flux through a closed surface is determined to be \(1.40 \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C},\) how much charge is enclosed by the surface?

In a follow-up experiment, a charge of \(+40 \mathrm{pC}\) was placed at the center of an artificial flower at the end of a \(30-\mathrm{cm}\) -long stem. Bees were observed to approach no closer than \(15 \mathrm{~cm}\) from the center of this flower before they flew away. This observation suggests that the smallest external electric field to which bees may be sensitive is closest to which of these values? A. \(2.4 \mathrm{~N} / \mathrm{C}\) B. \(16 \mathrm{~N} / \mathrm{C}\) C. \(2.7 \times 10^{-10} \mathrm{~N} / \mathrm{C}\) D. \(4.8 \times 10^{-10} \mathrm{~N} / \mathrm{C}\)

In a rectangular coordinate system, a positive point charge \(q=6.00 \mathrm{nC}\) is placed at the point \(x=+0.150 \mathrm{~m}, y=0,\) and an identical point charge is placed at \(x=-0.150 \mathrm{~m}, y=0 .\) Find the \(x\) and \(y\) components and the magnitude and direction of the electric field at the following points: (a) the origin; (b) \(x=0.300 \mathrm{~m}, y=0 ;\) (c) \(x=0.150 \mathrm{~m}, y=-0.400 \mathrm{~m} ;\) (d) \(x=0, y=0.200 \mathrm{~m}\)

Neurons are components of the nervous system of the body that transmit signals as electrical impulses travel along their length. These impulses propagate when charge suddenly rushes into and then out of a part of the neuron called an axon. Measurements have shown that, during the inflow part of this cycle, approximately \(5.6 \times 10^{11} \mathrm{Na}^{+}\) (sodium ions) per meter, each with charge \(+e,\) enter the axon. How many coulombs of charge enter a \(1.5 \mathrm{~cm}\) length of the axon during this process?

(a) An electron is moving east in a uniform electric field of \(1.50 \mathrm{~N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{~m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{~m}\) east of point \(A ?\) (b) \(\mathrm{A}\) proton is moving in the uniform electric field of part (a). At point A, the velocity of the proton is \(1.90 \times 10^{4} \mathrm{~m} / \mathrm{s}\), east. What is the speed of the proton at point \(B ?\)

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