Problem 54 - University Physics with Modern Physics

Question

A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) 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 \(\mathrm{cm}\) 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 \(\mathrm{cm}\) directly above its center.

Step-by-Step Solution

Answer

Straight wire: Apply integration, field is vertical upward. Circular wire: Symmetry results in vertical upward field.

Step 1: Understand the Problem

We have a charged wire of length 8.50 cm, with a linear charge density of +175 nC/m, lying straight on a table. We need to find the electric field at a point 6.00 cm above the midpoint of the wire. In the second part, after bending the wire into a circle, we need to find the electric field at the same point directly above the center of this circle.

Step 2: Set Up for Straight Wire

Initially, the wire is straight and lying on the table. To find the electric field at a point along the perpendicular bisector of the wire, we can use the formula for the electric field due to an infinite line of charge and modify it for a finite line segment.The line charge density \(\lambda = +175 \text{ nC/m} = 175 \times 10^{-9} \text{ C/m}\). The wire length is 8.5 cm which is equivalent to 0.085 meters.The electric field at a distance \(d = 6 \text{ cm} = 0.06 \text{ m}\) directly above the midpoint due to a finite line charge is calculated using integration over the length of the wire.

Step 3: Calculate Electric Field for Straight Wire

The formula for the electric field, E, due to a finite line of charge at its axis is given by:\[E = \frac{\lambda}{4\pi\varepsilon_0 d} \left(\frac{L/2}{\sqrt{(L/2)^2 + d^2}} + \frac{L/2}{\sqrt{(L/2)^2 + d^2}}\right),\]where \(\varepsilon_0\) is the permittivity of free space, approximately \(8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2\).Substituting the values \( \lambda = 175 \times 10^{-9} \text{ C/m} \), \(L = 0.085 \text{ m}\), and \(d = 0.06 \text{ m}\), calculate \(E\). The direction is vertically upward since the point is directly above the wire.

Step 4: Discuss the Direction for Straight Wire

Since the electric field is symmetrical about the midpoint of the wire and we are considering a point on the perpendicular bisector, the horizontal components from each infinitesimal charge segment cancel out. Thus, the electric field is vertically upward and along the vertical line above the midpoint of the wire.

Step 5: Set Up for Circular Wire

When the wire is bent into a circle, the net electric field at a point directly above the center of the circle is due to the symmetry of the charge distribution. The circle can be thought of having a radius \(r = \frac{L}{2\pi}\), where \(L = 0.085 \text{ m} \). We have a point at a height \( h = 0.06 \text{ m} \) above the center.

Step 6: Calculate Electric Field for Circular Wire

For a ring of charge, the vertical component of the electric field at an axial point is given by:\[E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{q \cdot h}{(r^2 + h^2)^{3/2}}.\]The total charge \(q = \lambda L = 175 \times 10^{-9} \times 0.085 \). Using these, calculate \(E\). The field direction is still vertically upward along the axis of the circle.

Step 7: Discuss the Direction for Circular Wire

The circular symmetry ensures that the electric field produced by each infinitesimal charge segment along the ring has a horizontal component that cancels out. Therefore, the entire contribution to the electric field is vertically upwards.

Key Concepts

Linear Charge Density

Linear charge density is a measure of how charge is distributed along a line, such as a wire. It is represented by the symbol \( \lambda \), and it's defined as the charge per unit length of the wire. In the exercise, the linear charge density \( \lambda \) is given as \(+175 \text{ nC/m}\). This means that every meter of the wire has 175 nanoCoulombs of charge spread uniformly along its length.
Understanding linear charge density is crucial when determining the electric field produced by a charged line, as it directly influences the strength of the field. Since field calculations often require integration over the length, knowing \( \lambda \) situates the problem in a more calculable framework.
When dealing with finite lines or specific shapes (like a circle), it's important to adjust the setup appropriately in terms of linear charge density to model the electric field accurately.

Permittivity of Free Space

Permittivity of free space, often denoted as \( \varepsilon_0 \), is a fundamental physical constant characterizing the ability of the vacuum to permit electric field lines. Its value is approximately \(8.85 \times 10^{-12} \, \text{C}^2/\text{N}\cdot\text{m}^2\).
This constant appears frequently in electrostatic equations, such as the ones used in finding the electric field due to different charge distributions. In the problem, \( \varepsilon_0 \) plays a key role in calculating the electric field generated by the wire. It occurs in the denominator of formulas to demonstrate its inverse relationship with force; namely, higher permittivity implies that a given charge produces a weaker field in the medium.
Understanding \( \varepsilon_0 \) helps in realizing how electric forces act in different environments, providing a baseline to compare dielectric materials.

Finite Line Charge

A finite line charge is a charge distribution along a limited-length straight line. Unlike an infinite line charge, which extends indefinitely, a finite line has clear boundaries. The exercise talks about how to compute the electric field at a point directly above the midpoint of a straight wire that represents a finite line charge.

When calculating the electric field from a finite line charge, the integration covers only the extent of the wire, differentiating it from calculations involving infinite lines where symmetry and simplification play larger roles. The resulting field not only depends on the distance from the line and the linear charge density but also requires considering the entire geometry due to the wire's limited length.

Knowing how a finite line charge operates allows you to solve problems involving real-world objects like wires and rods, giving a more practical perspective on electrical field distributions.