Question

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

(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

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).

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 \( \approx 8.85 \times 10^{-12} \text{ C}^2/\text{N} \cdot \text{m}^2 \).

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.

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.

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.

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.

Key concepts

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.