Short Answer 
The magnitude of the force acting on an alpha particle placed 4 cm from an infinite line of charge with a linear charge density of (10^{-7} , text{C/m}) is calculated by first determining the electric field, which is found to be (1.8 times 10^{3} , text{N/C}). The force on the alpha particle is then computed as (5.76 times 10^{-16} , text{N}), although the closest provided option is (7.2 times 10^{-15} , text{N}), suggesting discrepancies in the values.
Step 1: Understand the Problem
To find the magnitude of the force acting on an alpha particle, we need to comprehend the scenario presented. The alpha particle is placed at a distance of 4 cm from an infinite line of charge that has a linear charge density of (10^{-7} , text{C/m}). We will use the concept of electric fields created by lines of charge to solve for the force acting on it.
Step 2: Calculate the Electric Field
Using the formula for the electric field (E) produced by an infinite line of charge: (E = frac{2klambda}{r}), where (lambda) is the linear charge density and (k) is Coulomb’s constant, we can find the electric field. Substitute the values where (lambda = 10^{-7} , text{C/m}) and convert the distance from 4 cm to 0.04 m. The steps are as follows:
- Substitute (lambda) and (r) into the formula.
- Calculate (E) to find it as (1.8 times 10^{3} , text{N/C}).
Step 3: Compute the Force on the Alpha Particle
Now that we have the electric field, we can calculate the force (F) on the alpha particle using (F = qE). Here, the charge (q) of the alpha particle can be determined as (2e = 2 times (1.6 times 10^{-19} , text{C})). The steps to find (F) are:
- Determine (q) as (3.2 times 10^{-19} , text{C}).
- Multiply (q) and (E) to find (F = 5.76 times 10^{-16} , text{N}).
- Note that the closest option given is (7.2 times 10^{-15} , text{N}), indicating possible discrepancies in the values.