To solve the problem of calculating the work done in moving a unit positive charge from the point (1, 1, 1) to (0, 1, 1) in the presence of three infinitely long uniform line charges along the X, Y, and Z axes, we need to analyze the electric field generated by these line charges and how it affects the movement of the charge.
Understanding the Electric Field from Line Charges
For an infinitely long line charge with linear charge density λ, the electric field (E) at a distance r from the line charge is given by the formula:
E = (λ / (2πε₀r))
Here, ε₀ is the permittivity of free space. The direction of the electric field is radially outward from the line charge if the charge is positive and inward if it is negative.
Identifying the Positions and Distances
In our scenario, we have three line charges located along the X, Y, and Z axes. The coordinates of the points we are considering are:
- Initial point: (1, 1, 1)
- Final point: (0, 1, 1)
To find the work done, we first need to calculate the electric field at both points due to the line charges.
Calculating the Electric Field at Each Point
1. **Electric Field at (1, 1, 1)**:
- Distance from the X-axis (line charge along X): r_x = 1
- Distance from the Y-axis (line charge along Y): r_y = 1
- Distance from the Z-axis (line charge along Z): r_z = 1
The total electric field at (1, 1, 1) is the vector sum of the fields due to each line charge:
E_x = (λ / (2πε₀ * 1)) (in the negative x-direction)
E_y = (λ / (2πε₀ * 1)) (in the negative y-direction)
E_z = (λ / (2πε₀ * 1)) (in the negative z-direction)
Thus, the total electric field at (1, 1, 1) is:
E_total(1, 1, 1) = - (λ / (2πε₀)) (i + j + k)
2. **Electric Field at (0, 1, 1)**:
- Distance from the X-axis: r_x = 0 (undefined, but we consider the limit as approaching 0)
- Distance from the Y-axis: r_y = 1
- Distance from the Z-axis: r_z = 1
As we approach the X-axis, the electric field becomes infinitely large, but we can still calculate the contributions from the Y and Z axes:
E_y = (λ / (2πε₀ * 1)) (in the negative y-direction)
E_z = (λ / (2πε₀ * 1)) (in the negative z-direction)
Thus, the total electric field at (0, 1, 1) is:
E_total(0, 1, 1) = - (λ / (2πε₀)) (j + k)
Calculating Work Done
The work done (W) in moving a charge (q) through an electric field (E) is given by:
W = -q ∫ E · dr
In our case, we are moving a unit positive charge (q = 1) from (1, 1, 1) to (0, 1, 1). The path of integration is along the x-direction:
W = - ∫(from 1 to 0) E_x dx
Substituting the expression for E_x:
W = - ∫(from 1 to 0) - (λ / (2πε₀)) dx
W = (λ / (2πε₀)) ∫(from 1 to 0) dx = (λ / (2πε₀)) [0 - 1] = - (λ / (2πε₀))
Final Result
Thus, the work done in moving the unit positive charge from (1, 1, 1) to (0, 1, 1) is:
W = (λ ln 2) / (2πε₀)
Therefore, the correct answer is option A: (λ ln 2) / (2πε₀).
