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12 February, 06:24

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.)

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  1. 12 February, 10:22
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    E = 0.5*k*λ_o/x_o

    Step-by-step 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
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