91Ó°ÊÓ

Understanding the concept of charge per unit length is fundamental when dealing with electromagnetic fields around distributed charges. In essence, it provides a way to describe the density of electric charge along a linear object, like a wire. The charge per unit length, denoted by the Greek letter \( \lambda \), is calculated as the charge \( Q \) divided by the length of the object \( L \).

For a very long, straight wire, \( \lambda \) is assumed to be consistent across its entire length, enabling us to find the electric field created by the wire at various distances. In the provided exercise, the charge per unit length is given as \( 3.20 \times 10^{-10} \mathrm{C} / \mathrm{m} \), meaning that every meter of the wire contains this much charge.

Charges generate an electric field, which permeates the space around them, and knowing the charge per unit length allows us to calculate the electric field's magnitude at any point in space relative to the wire using the appropriate electric field formula.

The electric field formula connects the electric field \( E \) in the region surrounding a uniformly charged wire to the charge per unit length \( \lambda \) and the distance \( r \) from the wire. The formula is expressed as \( E = \frac {k \lambda} {r} \) where \( k \) is the Coulomb's constant \( (9.0 \times 10^{9} \mathrm{Nm}^{2}/\mathrm{C}^{2}) \) and represents the proportionality factor in Coulomb's law.

This equation reveals that the electric field \( E \) is directly proportional to \( \lambda \) and inversely proportional to \( r \), the distance from the charge. When deciphering the problem, by substituting the known values of \( \lambda \) and \( k \) into the electric field formula, one can solve for the desired value of \( r \) that yields a specific magnitude of the electric field, as demonstrated in our original exercise.

Solving for the radius, or distance from the wire, involves manipulating the electric field formula in order to isolate \( r \). In the context of the problem provided, the task is to find at what distance the electric field magnitude equals \( 2.50 \mathrm{~N} / \mathrm{C} \).

Once the known values are substituted into the formula, the next step is to perform algebraic operations to solve for \( r \). This step can be intuitive—if the electric field \( E \) is known along with \( k \) and \( \lambda \), you can simply divide both sides of the equation to leave \( r \) alone on one side. After dividing, the result is \( r = \frac {k \lambda} {E} \).

Completing the calculation with the provided numeric values gives the distance where the electric field magnitude is equal to \( 2.50 \mathrm{~N} / \mathrm{C} \). In the original exercise, the resulting radius is calculated to be 0.144 meters. It is crucial to follow such systematic steps to solve for \( r \) accurately, ensuring an understanding of the relationship between the electric field and its distance from the source charge.