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Chapter 18: Problem 54

A spherical surface completely surrounds a collection of charges.Find the electric flux through the surface if the collection consists of (a) a single \(+3.5 \times 10^{-6} \mathrm{C}\) charge, (b) a single \(-2.3 \times 10^{-6} \mathrm{C}\) charge, and \((\mathrm{c})\) both of the charges in (a) and (b).

Short Answer

Expert verified
(a) \(3.95 \times 10^5\), (b) \(-2.60 \times 10^5\), (c) \(1.36 \times 10^5\) N m虏/C.

Step by step solution

01

Understanding Gauss's Law

Gauss's Law relates the electric flux through a closed surface to the charge enclosed. Mathematically, it is given by:\[ \Phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0} \]where \(\Phi_E\) is the electric flux and \(Q_{\text{enclosed}}\) is the total charge inside the surface.
02

Solving for a Single Positive Charge

For the charge \(+3.5 \times 10^{-6} \text{ C}\), the flux through the spherical surface is:\[ \Phi_E = \frac{3.5 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 3.95 \times 10^5 \text{ N m}^2/\text{C} \]
03

Solving for a Single Negative Charge

For the charge \(-2.3 \times 10^{-6} \text{ C}\), the flux through the surface is:\[ \Phi_E = \frac{-2.3 \times 10^{-6}}{8.85 \times 10^{-12}} \approx -2.60 \times 10^5 \text{ N m}^2/\text{C} \]
04

Solving for Both Charges Combined

When both charges are inside, the total enclosed charge is:\[ Q_{\text{enclosed}} = 3.5 \times 10^{-6} + (-2.3 \times 10^{-6}) = 1.2 \times 10^{-6} \text{ C} \]The flux through the surface is:\[ \Phi_E = \frac{1.2 \times 10^{-6}}{8.85 \times 10^{-12}} \approx 1.36 \times 10^5 \text{ N m}^2/\text{C} \]

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Key Concepts

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

Electric Flux
Electric flux is a key concept in electromagnetism that helps us understand the strength and behavior of electric fields. Imagine you have a surface where an electric field runs through it, like sunshine streaming through a window. This passing of the electric field lines through a surface is what we refer to as electric flux.
  • The more electric field lines that penetrate through the surface, the greater the electric flux.
  • Mathematically, electric flux \(\Phi_E\) is given by Gauss's Law: \(\Phi_E = \frac{Q_{\text{enclosed}}}{\varepsilon_0}\), where \(Q_{\text{enclosed}}\) is the charge inside the surface and \(\varepsilon_0\) is the electric constant (permits the electric field to "flow").
Electric flux is measured in Newton-meters squared per Coulomb (N m\(^2\)/C). Understanding electric flux is essential in fields like electronics and physics, as it helps predict how charged particles influence each other through fields.
Spherical Surface
A spherical surface is a perfectly rounded boundary that can enclose charges within it. You can think of it like an invisible bubble surrounding a group of charged particles. This kind of geometric surface is significant when using Gauss's Law, as it provides symmetry that simplifies the calculations of electric flux.
  • A sphere's symmetry allows us to assume that the electric field magnitude is the same at every point on its surface, making calculations straightforward.
  • This assumption is true especially when dealing with uniformly charged distributions or a single point charge located at the center of the sphere.
Using a spherical surface is particularly useful when the distribution of charges creates uniform fields or when dealing with radially symmetric charge configurations. This helps break down complex electric field interactions into more manageable problems.
Enclosed Charge
The concept of enclosed charge is central to understanding how electric flux works with Gauss's Law. Enclosed charge refers to the total amount of electric charge contained within a closed surface, such as our spherical surface. It's like counting the marbles inside a jar.
  • In the context of Gauss's Law, the total electric flux through the surface depends solely on this enclosed charge, regardless of how the charge is distributed within.
  • For instance, if the charges are both positive and negative within the surface, their effects may cancel out partially or fully, impacting the net flux.
When solving problems using Gauss's Law, it is vital to calculate the net enclosed charge to accurately determine the electric flux through a closed surface. This concept helps simplify the analysis of complex charge environments, focusing on the total net result instead of individual charge positions or values.

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Most popular questions from this chapter

Four identical metal spheres have charges of \(q_{\mathrm{A}}=-8.0 \mu \mathrm{C}\) \(q_{\mathrm{B}}=-2.0 \mu \mathrm{C}, q_{C}=+5.0 \mu \mathrm{C},\) and \(q_{\mathrm{D}}=+12.0 \mu \mathrm{C}\) (a) Two of the spheres are brought together so they touch, and then they are separated. spheres are brought together so they touch, and then they are separated. Which spheres are they, if the final charge on each one is \(+5.0 \mu \mathrm{C} ?\) (b) In a similar manncr, which three spheres are brought together and then separated, if the final charge on each of the three is \(+3.0 \mu \mathrm{C} ?\) (c) The final charge on each of the three separated spheres in part (b) is \(+3.0 \mu \mathrm{C}\) . How many clectrons would have to be addcd to onc of these spheres to make it electrically neutral?

The drawing shows an edge-on view of two planar surfaces that intersect and are mutually perpendicular. Surface 1 has an area of \(1.7 \mathrm{m}^{2},\) while surface 2 has an area of 3.2 \(\mathrm{m}^{2}\) The clectric ficld \(\overrightarrow{\mathbf{E}}\) in the drawing is uniform and has a drawing is uniform and has a magnitude of 250 \(\mathrm{NC}\) . Find the magnitude of the electric flux through (a) surface 1 and \((\mathrm{b})\) surface 2 .

Multiple-Concept Example 9 illustrates the concepts employed in this problem. A small object, which has a charge q 7.5 C and mass m 9.0 105 kg, is placed in a constant electric field. Starting from rest, the object accelerates to a speed of 2.0 103 m/s in a time of 0.96 s. Determine the magnitude of the electric field.

ssm At a distance \(r_{1}\) from a point charge, the magnitude of the electric field created by the charge is 248 \(\mathrm{N} / \mathrm{C}\) . At a distance \(r_{2}\) from the charge, the ficld has a magnitude of 132 \(\mathrm{N} / \mathrm{C}\) . Find the ratio \(r_{2} / r_{1}\) .

A charge of \(-3.00 \mu \mathrm{C}\) is fixed at the center of a compass. Two additional charges are fixed on the circle of the compass, which has a radius of \(0.100 \mathrm{~m}\). The charges on the circle are \(-4.00 \mu \mathrm{C}\) at the position due north and \(+5.00 \mu \mathrm{C}\) at the position due east. What are the magnitude and direction of the net electrostatic force acting on the charge at the center? Specify the direction relative to due east.
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