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Ann [662]
4 years ago
7

A continuous line of charge lies along the x axis, extending from x = +x0 to positive infinity. The line carries positive charge

with a uniform linear charge density λ0. (a) What is the magnitude of the electric field at the origin? (Use the following as necessary: λ0, x0, and ke.) E = (b) What is the direction of the electric field at the origin? +x −x +y −y +z −z
Physics
1 answer:
yawa3891 [41]4 years ago
7 0

Answer:

 E = k λ₀ / x₀, the field is in thenegative direction of the x axis (-x)

Explanation:

In this problem the electric field of a line of charge is requested, the expression for the electric field is

          E = k ∫ dq / r²

where k is the Coulomb constant that you are worth 9 10⁹ N m²/C², that the charge and r the distance to the point of interest, in this case it is the origin (x = 0)

let's use the definite linear density

        λ₀ = dq / dx

        dq = λ₀ dx

we replace and integrate

       E = k λ₀ ∫ dx / x²

       E = k λ₀ ( -1 / x)

we evaluate the integral from the lower limit of load x = x₀ to the upper limit x = ∞

       E = - k λ₀ (1 /∞ - 1 / x₀)

       E = k λ₀ / x₀

     

as the field is positive the direction is away from the charges, so it is in the negative direction of the x axis (-x)

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