Question

A total of \(3.05 \cdot 10^{6}\) electrons are placed on an initially uncharged wire of length \(1.33 \mathrm{~m}\) a) What is the magnitude of the electric field a perpendicular distance of \(0.401 \mathrm{~m}\) away from the midpoint of the wire? b) What is the magnitude of the acceleration of a proton placed at that point in space? c) In which direction does the electric field force point in this case?

Short answer

Also, determine the direction of the electric field. Answer: The acceleration of the proton at a distance of 0.401 m from the midpoint of the charged wire is approximately \(7.97 \times 10^{11} \mathrm{m/s^2}\). The direction of the electric field is attractive and points toward the midpoint of the charged wire.

Step-by-step solution

Find the charge on the wire

To find the total charge on the wire, multiply the charge of a single electron by the total number of electrons: \(q_{total} = n \cdot e\), where \(n = 3.05 \cdot 10^6\) is the number of electrons and \(e = 1.60 \cdot 10^{-19} \mathrm{C}\) is the charge of a single electron.

Calculate the linear charge density

Divide the total charge by the length of the wire to find the linear charge density \(\lambda\): \(\lambda = \frac{q_{total}}{L}\), where \(L = 1.33 \mathrm{m}\) is the length of the wire.

Calculate the electric field at the given point

To find the magnitude of the electric field, use the formula \(E = \frac{1}{4\pi\epsilon_0} \frac{\lambda}{R}\), where \(R = 0.401 \mathrm{m}\) is the distance from the midpoint of the wire, and \(ε_0 = 8.85 \cdot 10^{-12} \mathrm{C^2/N \cdot m^2}\) is the permittivity of free space.

Calculate the force experienced by the proton

To find the electric force experienced by the proton at the given point, use the formula \(F = qE\), where \(q\) is the charge of the proton, which is equal to \(1.60 \cdot 10^{-19} \mathrm{C}\) (the same as an electron, but positive).

Calculate the acceleration of the proton

To find the acceleration of the proton at the given point, use Newton's second law: \(a = \frac{F}{m}\), where \(m = 1.67 \cdot 10^{-27} \mathrm{kg}\) is the mass of the proton.

Determine the direction of the electric field force

Since the wire is negatively charged and a proton is positively charged, the electric field force will be attractive and point toward the midpoint of the charged wire.