Question

A line of charge starts at x = x0, where x0 is positive, and extends along the x-axis to positive infinity. If the linear charge density is given by λ = λ0 x0/x, where λ0 is a positive constant, find the electric field at the origin. (ˆı denotes the unit vector in the positive x direction.)

Answer 1

E = 0.5*k*λ_o/x_o

Explanation:

Given:

- A line of charge with a variable charge density:

λ = λ_o*x_o/x

- The line of charge starts from x = x_o to +infinity.

Find:

Find the electric field at the origin.

Solution:

- We will first develop an expression of a differential section of the line that causes an electric field at the origin as follows:

dE = k*λ.dx / x^2

- Remember that λ is a function of x as given in the expression. So will plug the expression of charge density into the differential form:

dE = (k*λ_o*x_o/x^3). dx

- Now, we will sum up all the differential elements of the line of charge. We will integrate the expression from x = x_o to x =+inf.

E = k*λ_o*x_o integral ( 1 / x^3 ) .dx

E = -0.5*k*λ_o*x_o ( 1 / x^2)

Evaluate the limits:

E = -0.5*k*λ_o*x_o ( 0 - 1/x_o^2)

E = 0.5*k*λ_o/x_o