Question

A long conducting wire with charge distribution \(\lambda\) and radius \(r\) produces an electric field of \(2.73 \mathrm{~N} / \mathrm{C}\) just outside its surface. What is the magnitude of the electric field just outside the surface of another wire with charge distribution \(0.810 \lambda\) and radius \(6.50 r ?\)

Short answer

Answer: The magnitude of the electric field just outside the surface of the second wire is 0.339 N/C.

Step-by-step solution

Recall Gauss's Law for a long conducting wire

Gauss's law states that the electric flux through a closed Gaussian surface is equal to the total charge enclosed by the surface divided by the permittivity of free space. For a long conducting wire with a cylindrical Gaussian surface, this can be expressed as: \(E \cdot 2\pi r L = \frac{\lambda L}{\epsilon_0}\) where \(E\) is the electric field, \(L\) is the length of the wire, \(r\) is the radius of the wire, \(\lambda\) is the charge distribution and \(\epsilon_0\) is the permittivity of free space.

Calculate the electric field for the first wire

Since we know the electric field just outside the first wire's surface, we can plug the given value of \(E\) and rearrange the Gauss's law equation to solve for the charge distribution \(\lambda\): \(\lambda = 2\pi r E \epsilon_0\) Substituting the given values, we get: \(\lambda = 2\pi r(2.73 N/C) \epsilon_0\)

Calculate the charge distribution of the second wire

We are given that the charge distribution of the second wire is \(0.810\lambda\). Therefore, multiply the previously calculated value for \(\lambda\) by \(0.810\) to obtain the charge distribution \(\lambda_2\) for the second wire: \(\lambda_2 = 0.810 \cdot 2\pi r(2.73 N/C) \epsilon_0\)

Calculate the electric field just outside the surface of the second wire using Gauss's Law

Now, we use Gauss's law again for the second wire with charge distribution \(\lambda_2\) and radius \(6.50r\). We have: \(E_2 \cdot 2\pi (6.50r) L = \frac{\lambda_2 L}{\epsilon_0}\) Rearrange the equation to solve for \(E_2\): \(E_2 = \frac{\lambda_2}{2\pi (6.50r) \epsilon_0}\)

Substitute the values for \(\lambda_2\) and \(r\) and solve for \(E_2\)

Now, plug in \(\lambda_2 = 0.810 \cdot 2\pi r(2.73 N/C) \epsilon_0\) and \(6.50r\) from the problem: \(E_2 = \frac{0.810 \cdot 2\pi r(2.73 N/C) \epsilon_0}{2\pi (6.50r) \epsilon_0}\) Simplify and solve for \(E_2\): \(E_2 = \frac{0.810(2.73 N/C)}{6.50} = 0.339 N/C\) Thus, the magnitude of the electric field just outside the surface of the second wire is \(0.339 N/C\).

Key concepts

Electric Field

The electric field is a vector quantity used to describe the influence of electric charges in the space around them. It is defined as the force per unit charge experienced by a positive test charge placed in the field. The direction of the electric field is the direction of the force on a positive charge.

In essence, the electric field represents how charged objects interact with one another without touching. For example, when calculating the electric field produced just outside the surface of a conducting wire, the field can be visualized as radiating outward from the wire, exerting force on charges or objects nearby.

Charge Distribution

Charge distribution refers to how electric charge is spread out in space. In the context of conducting wires, the charge can be uniformly distributed along the length of the wire. This is often characterized by a charge per unit length (linear charge density \( \lambda \) ).

Understanding this concept is vital for applying Gauss's Law because the electric field generated by the wire depends on its charge distribution. The type of charge distribution affects the resulting electric field pattern, which can be symmetrical for uniform distributions or more complex for non-uniform distributions.

Conducting Wire

A conducting wire in physics is typically a wire made of a material that allows electrons to move freely, such as copper or aluminum.

These wires are often studied because they are assumed to have a uniform positive charge distribution along their length. This simplifies calculations involving electric fields generated by the wires using Gausss Law. In the context of Gauss's Law, the conducting wire's surface can act as a Gaussian surface to help calculate the electric field generated at points outside the wire.

Permittivity of Free Space

The permittivity of free space, denoted by \( \epsilon_0 \), is a fundamental constant in electromagnetism. It describes how much resistance is encountered when forming an electric field in a vacuum.

Its value affects the strength of the electric field generated by a given charge distribution, as seen in the formula derived from Gauss's Law. Specifically, \( \epsilon_0 \) is used in calculating the electric fields around a charge distribution, as a higher permittivity means the medium opposes the electric field more, leading to a weaker field for the same amount of charge.