A very long uniformly charged, thin rod is oriented along x-axis with its one end at origin and it extends along +x-axis to infinity. The charge per unit length of the rod is +λ. Calculate the electric field vector at x = – a on x-axis.

To calculate the electric field vector at x = -a on the x-axis due to the uniformly charged rod, we can use the principle of superposition.

Step 1: Divide the rod into small segments of length dx.

Step 2: Consider a small segment of the rod located at a distance x from the origin. The charge dq of this segment can be calculated as dq = λ*dx, where λ is the charge per unit length of the rod.

Step 3: Calculate the electric field vector dE due to this small segment at point P(x = -a, y = 0, z = 0) using Coulomb's law. The magnitude of dE can be calculated as dE = k * dq / r^2, where k is the electrostatic constant and r is the distance between the small segment and point P.

Step 4: The x-component of the electric field vector dE_x due to this small segment can be calculated as dE_x = dE * cosθ, where θ is the angle between the x-axis and the line connecting the small segment to point P. Since the rod is oriented along the x-axis, θ will be 0 degrees. Therefore, cosθ = 1.

Step 5: Integrate the x-component of the electric field vector dE_x over the entire length of the rod to find the total electric field vector at point P.

The integral can be written as E_x = ∫dE_x = ∫dE * cosθ = ∫(k * dq / r^2) * 1

Step 6: Substitute the values of dq, r, and cosθ into the integral.

E_x = ∫(k * λ * dx / r^2)

Step 7: Calculate the limits of integration. Since the rod extends along the positive x-axis to infinity, the limits of integration will be from 0 to infinity.

E_x = ∫(k * λ * dx / r^2) from 0 to infinity

Step 8: Evaluate the integral.

E_x = k * λ * ∫(dx / r^2) from 0 to infinity

Step 9: Calculate the value of r at x = -a. Since r is the distance between the small segment and point P, r = a + x.

Step 10: Substitute the value of r into the integral.

E_x = k * λ * ∫(dx / (a + x)^2) from 0 to infinity

Step 11: Evaluate the integral using the limits of integration.

E_x = k * λ * [(-1 / (a + x))] from 0 to infinity

Step 12: Substitute the limits of integration into the expression.

E_x = k * λ * [(-1 / (a + infinity)) - (-1 / (a + 0))]

Step 13: Simplify the expression.

E_x = k * λ * [(-1 / infinity) - (-1 / a)]

Step 14: Since 1 / infinity is equal to 0, the expression becomes:

E_x = k * λ * (0 - (-1 / a))

Step 15: Simplify further.

E_x = k * λ / a

Therefore, the electric field vector at x = -a on the x-axis due to the uniformly charged rod is E_x = k * λ / a, where k is the electrostatic constant and λ is the charge per unit length of the rod.