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The electric field 0.400 m from a very long uniform line of charge is 840 N/C. How much charge is contained in a 2.00-cm section of the line?

Short Answer

Expert verified
The charge in a 2.00-cm section is 3.76 × 10^{-10} C.

Step by step solution

01

Understand Gauss’s Law

Gauss's Law relates the electric field caused by a charged object to the charge enclosed by a Gaussian surface. For an infinite line of charge, the electric field at a distance r is given by:\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]where \( E \) is the electric field, \( \lambda \) is the linear charge density, \( \varepsilon_0 \) is the vacuum permittivity, and \( r \) is the distance from the line of charge.
02

Apply Gauss’s Law to Find Linear Charge Density

Rearrange Gauss’s Law to solve for the linear charge density \( \lambda \):\[ \lambda = 2\pi\varepsilon_0 r E \]Substitute the known values: \( r = 0.400 \) m and \( E = 840 \) N/C. The value of \( \varepsilon_0 = 8.85 \times 10^{-12} \) C²/(N·m²).\[ \lambda = 2\pi(8.85 \times 10^{-12})(0.400)(840) \]
03

Calculate Linear Charge Density

Perform the calculation:\[ \lambda = 2\pi(8.85 \times 10^{-12})(0.400)(840) \approx 1.88 \times 10^{-8} \text{ C/m} \]This is the charge per meter along the line of charge.
04

Calculate Total Charge in a 2-cm Section

The charge in a 2.00-cm section of the line is given by:\[ Q = \lambda \times \text{length} \]Convert the length from cm to m: 2.00 cm = 0.0200 m. Substitute the values:\[ Q = (1.88 \times 10^{-8} \text{ C/m}) \times 0.0200 \text{ m} \]
05

Calculate and Provide the Charge

Perform the calculation:\[ Q = 1.88 \times 10^{-8} \times 0.0200 = 3.76 \times 10^{-10} \text{ C} \]Thus, 3.76 \times 10^{-10} C of charge is contained in a 2.00-cm section of the line.

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in physics that describes the force experienced by a charge in the vicinity of other charged objects. It is a vector field, meaning it has both magnitude and direction. When we talk about the electric field (\( E \)) generated by a source, we refer to the force per unit charge that a positive test charge would feel within this field's influence. An electric field is often described using field lines that indicate the direction and strength of the field at various points.
If you imagine a charged line, like the one in the exercise, the electric field it produces impacts the space around it. For a very long line of charge, the electric field is determined by the linear charge density and diminishes with distance, providing a useful abstract for analyzing uniform charge distributions. The formula for the electric field from an infinite line of charge is:\[ E = \frac{\lambda}{2\pi\varepsilon_0 r} \]where:
  • \( E \) is the electric field strength,
  • \( \lambda \) is the linear charge density,
  • \( \varepsilon_0 \) is the permittivity of free space (a constant), and
  • \( r \) is the radial distance from the line.
Linear Charge Density
Linear charge density, denoted by \( \lambda \), describes how charge is distributed along a line. It is the amount of charge per unit length, usually expressed in coulombs per meter (C/m). This concept is especially useful when dealing with problems involving infinite or very long charged objects, like wires or rods, where the charge is symmetrically spread along the line.
In the original problem, calculating linear charge density involves rearranging Gauss’s Law to express it in terms of known quantities: the electric field, the radial distance from the line, and the permittivity of free space. By deriving:\[ \lambda = 2\pi\varepsilon_0 r E \]you can determine the specific charge density at a given distance.
This parameter helps us understand the distribution of charge along the line and serves as a stepping stone for calculations involving the total charge over specific intervals—like the 2-cm segment mentioned in the exercise.
Charge Distribution
Charge distribution refers to how electric charges are spread over a given object or space. This can occur in several forms:
  • Linear, as seen in long wires where charge per unit length is considered (\( \lambda \)).
  • Surface, where charges are spread across a plane, calculated as charge per unit area.
  • Volume, which deals with charges distributed throughout a volume, calculated as charge per unit volume.

In the specific scenario of an infinitely long line, we focus on linear charge distribution, which simplifies many calculations due to the symmetry involved. By understanding the charge distribution, we can apply Gauss's Law to solve for the electric field and further determine how much charge is contained in specific parts of the line.
As in our exercise, once the linear charge density is known, calculating the total charge in any section becomes straightforward: multiply the linear charge density by the section's length. This clear-cut relationship makes it easy to understand how charges are arranged in various physical contexts.

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Most popular questions from this chapter

A 6.20-\(\mu\)C point charge is at the center of a cube with sides of length 0.500 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.

A solid conducting sphere with radius \(R\) carries a positive total charge \(Q\). The sphere is surrounded by an insulating shell with inner radius \(R\) and outer radius 2\(R\). The insulating shell has a uniform charge density \(\rho\). (a) Find the value of \(\rho\) so that the net charge of the entire system is zero. (b) If \(\rho\) has the value found in part (a), find the electric field \(\overrightarrow{E}\) (magnitude and direction) in each of the regions 0 \(< r < R, R < r < 2R\), and \(r > 2R\). Graph the radial component of \(\overrightarrow{E}\) as a function of r. (c) As a general rule, the electric field is discontinuous only at locations where there is a thin sheet of charge. Explain how your results in part (b) agree with this rule.

A very large, horizontal, nonconducting sheet of charge has uniform charge per unit area \(\sigma =\) 5.00 \(\times\) 10\(^{-6}\) C/m\(^2\). (a) A small sphere of mass \(m =\) 8.00 \(\times\) 10\(^{-6}\) kg and charge \(q\) is placed 3.00 cm above the sheet of charge and then released from rest. (a) If the sphere is to remain motionless when it is released, what must be the value of \(q\)? (b) What is \(q\) if the sphere is released 1.50 cm above the sheet?

A cube has sides of length \(L =\) 0.300 m. One corner is at the origin (Fig. E22.6). The nonuniform electric field is given by \(\overrightarrow{E} =\) (-5.00 N/C \(\cdot\) m)\(x\hat{\imath}\) + (3.00 N/C \(\cdot\) m)\(z \hat{k}\). (a) Find the electric flux through each of the six cube faces \(S_1, S_2, S_3, S_4, S_5\), and \(S_6\) . (b) Find the total electric charge inside the cube.

A very long, solid cylinder with radius \(R\) has positive charge uniformly distributed throughout it, with charge per unit volume \(\rho\). (a) Derive the expression for the electric field inside the volume at a distance \(r\) from the axis of the cylinder in terms of the charge density \(\rho\). (b) What is the electric field at a point outside the volume in terms of the charge per unit length \(\lambda\) in the cylinder? (c) Compare the answers to parts (a) and (b) for \(r = R\). (d) Graph the electric-field magnitude as a function of r from \(r = 0\) to \(r = 3R\).

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