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Chapter 21: Problem 52

A straight, nonconducting plastic wire 8.50 cm long carries a charge density of \(+\)175 nC\(/\)m distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 cm directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 cm directly above its center.

Short Answer

Expert verified
(a) The magnitude of the electric field is approximately 1.65 N/C and it points radially outward. (b) The magnitude is about 1.75 N/C and it points vertically upward.

Step by step solution

01

Calculate the linear charge density

The linear charge density \( \lambda \) is given as \( +175 \text{ nC/m} \) or \( +175 \times 10^{-9} \text{ C/m} \). As the charge is uniformly distributed, this remains constant for both parts (a) and (b).
02

Determine the electric field at a point above a straight wire

The electric field due to a uniformly charged straight wire at a point along the perpendicular bisector is given by the formula: \[ E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2\lambda}{r} \cdot \left( \frac{L}{\sqrt{L^2 + 4r^2}} \right) \]Where \( L \) is the length of the wire (8.50 cm or 0.085 m), \( r \) is the perpendicular distance from the wire (6.00 cm or 0.060 m), and \( \varepsilon_0 \) is the permittivity of free space \( \approx 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).
03

Plug values to find electric field for part (a)

Plugging the values into the formula from Step 2:\[ E = \frac{1}{4\pi \times 8.85 \times 10^{-12}} \cdot \frac{2 \times 175 \times 10^{-9}}{0.060} \cdot \left( \frac{0.085}{\sqrt{0.085^2 + 4 \times 0.060^2}} \right) \]Calculate the above expression to find the value of the electric field's magnitude. The direction of the electric field is radially outward from the wire.
04

Determine the electric field at a point above a circular wire

For part (b), when the wire is bent into a circle, the electric field at a point above the center of the circle is due to the symmetry of the circle. Here, the net electric field along the horizontal directions cancels out, and only the vertical component will remain.
05

Use formula for the electric field of a ring

The electric field above the center of a ring is given by:\[ E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Qz}{(z^2 + R^2)^{3/2}} \]Where \( Q \) is the total charge on the ring, \( z = 0.060 \) m is the height above the center, and \( R = \frac{L}{2\pi} \) is the radius of the ring, with the wire now being in the shape of a circle with the same length \( L = 0.085 \) m.
06

Calculate the electric field for part (b)

First find the radius \( R \) of the circle: \[ R = \frac{0.085}{2\pi} \approx 0.0135 \text{ m} \]Total charge \( Q = \lambda \times L = 175 \times 10^{-9} \times 0.085 \text{ C} \). Now calculate the electric field using the formula from Step 5:\[ E = \frac{1}{4\pi \times 8.85 \times 10^{-12}} \cdot \frac{(175 \times 10^{-9} \times 0.085) \times 0.060}{(0.060^2 + 0.0135^2)^{3/2}} \]The direction of this electric field would be vertically upward.

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Key Concepts

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

Understanding Linear Charge Density
Linear charge density, represented by \( \lambda \), is a measure of the amount of electric charge distributed along a line, such as a straight wire or a ring. In this exercise, the wire has a charge density of \(+175 \text{ nC/m}\), meaning that each meter of the wire contains \(175\times 10^{-9} \text{ C}\) of charge. This concept is important because the electric field produced by the wire depends directly on the charge density.
  • Units: Usually expressed in terms of charge per unit length, like nanocoulombs per meter \(\text{ nC/m}\).
  • Uniform distribution: The charge must be distributed evenly along the wire or circle.
This simplicity aids in calculating electric fields because it allows us to treat each small segment of the wire's length as having the same contribution to the field.
The Role of Permittivity of Free Space
Permittivity of free space, denoted as \( \varepsilon_0 \), is a fundamental physical constant that describes how electric fields interact with a vacuum. The value is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \). This parameter appears in the equations used to calculate electric fields.
  • Significance: It affects the strength of the electric field produced by a charged object.
  • In equations: Used in formulas such as \( E = \frac{1}{4\pi\varepsilon_0} \cdots \).
Understanding this constant is crucial when calculating the electric field's magnitude, as it scales the effect of the charge density and geometry.
Calculating Electric Field Due to a Wire
When we need to find the electric field created by a straight wire at a particular point, we use the linear charge density and geometric setup. The electric field due to a straight wire has a formula based on symmetry and calculus: \[E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2\lambda}{r} \cdot \left( \frac{L}{\sqrt{L^2 + 4r^2}} \right)\]
  • \( L \): Length of the wire.
  • \( r \): Distance from the wire.
  • \( \lambda \): Charge density.
The field is strongest directly perpendicular to the wire and weakens as we move further away. Its direction is radially outward from the wire, assuming positive charge density.
Electric Field Due to a Ring
When the straight wire is bent into a ring, the way the electric field behaves changes due to circular symmetry. Now, the field is outward along the axis of the ring's circular plane.
The formula for the electric field at a point directly above the center of the ring is:\[E = \frac{1}{4\pi \varepsilon_0} \cdot \frac{Qz}{(z^2 + R^2)^{3/2}}\]
  • \( Q \): Total charge on the ring.
  • \( z \): Height above the center of the ring.
  • \( R \): Radius of the ring, found via \( R = \frac{L}{2\pi}\).
Symmetry simplifies the calculations because contributions to the electric field from different parts of the ring cancel in the horizontal direction, leaving only the vertical component.
Exploring Symmetry in Electric Fields
Symmetry plays a crucial role in simplifying calculations of electric fields, particularly when dealing with geometrically symmetric objects like a straight wire or a ring. In this exercise:
  • For the straight wire: The electric field is symmetric around the axis perpendicular to the wire, meaning it is radial and depends inversely on distance.
  • For the ring: When the wire is bent into a circle, symmetry allows us to focus on the vertical component of the electric field, as horizontal components cancel out.
Symmetry helps reduce complexity, making it easier to analyze the effect of geometry on the electric field's magnitude and direction. Recognizing symmetry can accelerate the problem-solving process significantly.

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Most popular questions from this chapter

Two identical spheres are each attached to silk threads of length \(L =\) 0.500 m and hung from a common point (Fig. P21.62). Each sphere has mass \(m =\) 8.00 g. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_1\) , and the other a different positive charge \(q_2\) ; this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta = 20.0^\circ\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the given information, what can you say about the magnitudes of \(q_1\) and \(q_2\)? Explain. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of 30.0\(^\circ\) with the vertical. Determine the original charges. (\(Hint\): The total charge on the pair of spheres is conserved.)

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.4 \(\times \space 10^{-8}\) g. The drops will leave the nozzle and travel toward the paper at 50 m\(/\)s, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2.0 cm long, where there is a uniform vertical electric field with magnitude 8.0 \(\times \space 10^4 \space N/C\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 \(m/s\), what \(q\) value is needed to achieve the same 0.30-mm deflection?

If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

A point charge is at the origin. With this point charge as the source point, what is the unit vector \(\hat{r}\) in the direction of the field point (a) at \(x = 0, \space y = -\)1.35 m; (b) at \(x =\) 12.0 cm, \(y =\) 12.0 cm; (c) at \(x = -\)1.10 m, \(y =\) 2.60 m ? Express your results in terms of the unit vectors \(\hat{\imath}\) and \(\hat{\jmath}\).

In a follow-up experiment, a charge of \(+40\) pC was placed at the center of an artificial flower at the end of a 30-cm long stem. Bees were observed to approach no closer than 15 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) \(2.4 \space N/C; (b) 16 \space N/C; (c) 2.7 \times 10^{-10} \space N/C; (d) 4.8 \times 10^{-10} \space N/C.\)
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