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Chapter 17: Problem 74

\(\bullet$$\bullet\) A \(9.60-\mu \mathrm{C}\) point charge is at the center of a cube with sides of length 0.500 \(\mathrm{m}\) . (a) What is the electric flux through one of the six faces of the cube? (b) How would your answer to part (a) change if the sides were 0.250 m long? Explain.

Short Answer

Expert verified
The electric flux through one face is \( 1.81 \times 10^5 \frac{\text{Vm}}{\text{As}} \). Changing the side length doesn't change the flux.

Step by step solution

01

Understanding the Problem

We need to find the electric flux through one face of a cube that contains a point charge at its center. Electric flux through a closed surface enclosing a charge is calculated using Gauss's Law: \( \Phi = \frac{Q}{\varepsilon_0} \), where \( Q \) is the charge enclosed, and \( \varepsilon_0 \) is the permittivity of free space (\( \varepsilon_0 = 8.854 \times 10^{-12} \frac{\text{As}}{\text{Vm}} \)). Since the cube has six faces, the flux through one face is one-sixth of the total flux.
02

Calculating Total Electric Flux Using Gauss's Law

We'll calculate the total electric flux through the cube using Gauss's Law. The charge \( Q \) is given as \( 9.60 \times 10^{-6} \) C, and the permittivity of free space is \( 8.854 \times 10^{-12} \frac{\text{As}}{\text{Vm}} \). Therefore, the total flux is \( \Phi = \frac{9.60 \times 10^{-6}}{8.854 \times 10^{-12}} \).
03

Computing Total Electric Flux

Compute the total electric flux using the values: \( \Phi = \frac{9.60 \times 10^{-6}}{8.854 \times 10^{-12}} \approx 1.084 \times 10^6 \frac{\text{Vm}}{\text{As}} \).
04

Finding Electric Flux Through One Face

Since the cube has six faces, the flux through one face is \( \frac{1.084 \times 10^6}{6} \approx 1.81 \times 10^5 \frac{\text{Vm}}{\text{As}} \).
05

Addressing the Change in Side Length

According to Gauss's Law, the electric flux through a closed surface only depends on the charge enclosed, and not on the size of the surface. Therefore, if the sides of the cube are changed to 0.250 m, the total flux, and hence the flux through each face, remains the same.
06

Final Answer

The electric flux through one face of the cube is approximately \( 1.81 \times 10^5 \frac{\text{Vm}}{\text{As}} \), and changing the side length doesn't affect this value.

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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 crucial concept in understanding electric fields. It essentially measures how much of the electric field passes through a given surface. Imagine electric field lines passing through a surface; electric flux quantifies these interactions.

The formula to calculate electric flux, using Gauss's Law, is \( \Phi = \frac{Q}{\varepsilon_0} \). This indicates that electric flux is related to the charge \( Q \) enclosed by the surface and the permittivity of free space \( \varepsilon_0 \).

Key points to remember:
  • Electric flux depends on the charge enclosed, not the size of the surface.
  • The concept is central in Gauss's Law, connecting it to electric fields.
  • For symmetric shapes, it is often feasible to evenly distribute flux across different faces of the shape.
Permittivity of Free Space
The permittivity of free space, often denoted as \( \varepsilon_0 \), is a fundamental constant that plays a critical role in electrostatics. It factors into Gauss's Law and the calculation of electric flux.

Mathematically, \( \varepsilon_0 \) is approximately equal to \( 8.854 \times 10^{-12} \frac{\text{As}}{\text{Vm}} \).
  • It links charge with the resultant electric field.
  • It defines the capacity of the vacuum to permit electric field lines.
  • A higher \( \varepsilon_0 \) means weaker interactions between charges.
Understanding \( \varepsilon_0 \) helps us comprehend how electric fields propagate through space and is integral in various equations of electromagnetism.
Point Charge
A point charge is an idealized model of a charged particle. It is used extensively in physics to simplify the analysis of electric fields.

In this idealization:
  • The charge is located at a single point in space.
  • We assume it has negligible size.
  • It helps in studying the electric field and force around compact charged objects.
Using a point charge makes it easier to apply Gauss's Law, especially when trying to find electric flux through closed surfaces.

This exercise involved a point charge at the center of a cube, showcasing its central importance in both teaching and application of electrostatics.
Closed Surface
A closed surface is an essential concept in the application of Gauss's Law. It is any surface that completely encloses a volume without any gaps or boundaries left open. Examples include cubes, spheres, and cylinders.

When we use Gauss's Law:
  • The closed surface must enclose the charge for the law to be applicable.
  • The electric flux through this surface is entirely determined by the charge enclosed.
  • A closed surface helps isolate a volume of space to calculate field interactions.
In our exercise, the cube acts as the closed surface enclosing the point charge. This containment allows for the straightforward calculation of electric flux, demonstrating why closed surfaces are vital in electromagnetic theory.

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Most popular questions from this chapter

\(\bullet\) \(\bullet\) Sketch electric field lines in the vicinity of two charges, \(Q\) and \(-4 Q,\) located a small distance apart on the \(x\) -axis.

\(\bullet$$\bullet\) Two small aluminum spheres, each having mass 0.0250 \(\mathrm{kg}\) , are separated by 80.0 \(\mathrm{cm}\) . (a) How many electrons does each sphere contain? (The atomic mass of aluminum is 26.982 \(\mathrm{g} / \mathrm{mol}\) , and its atomic number is \(13 .\) (b) How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude \(1.00 \times 10^{4} \mathrm{N}\) (roughly 1 ton)? Assume that the spheres may be treated as point charges. (c) What fraction of all the elec- trons in each sphere does this represent?

\(\bullet$$\bullet\) An electron is released from rest in a uniform electric field. The electron accelerates vertically upward, traveling 4.50 \(\mathrm{m}\) in the first 3.00\(\mu\) s after it is released. (a) What are the magnitude and direction of the electric field? (b) Are we justified in ignor- ing the effects of gravity? Justify your answer quantitatively.

\(\bullet\) A small object carrying a charge of \(-8.00 \mathrm{nC}\) is acted upon by a downward force of 20.0 \(\mathrm{nN}\) when placed at a certain point in an electric field. (a) What are the magnitude and direction of the electric field at the point in question? (b) What would be the magnitude and direction of the force acting on a proton placed at this same point in the electric field'?

\(\bullet$$\bullet\) A small 12.3 g plastic ball is tied to a very light 28.6 \(\mathrm{cm}\) string that is attached to the vertical wall of a room. (See Figure \(17.58 . ) \mathrm{A}\) uniform horizontal electric field exists in this room. When the ball has been given an excess charge of \(-1.11 \mu \mathrm{C},\) you observe that it remains suspended, with the string making an angle of \(17.4^{\circ}\) with the wall. Find the magnitude and direction of the electric field in the room.
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