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An infinite line of charge with charge density \lambda_1 = 4.6 \mu C/cm is aligned with the...

Question:

An infinite line of charge with charge density {eq}\lambda_1 = 4.6 \mu C/cm {/eq} is aligned with the {eq}y {/eq}-axis as shown.

What is {eq}E_x(P) {/eq}, the value of the {eq}x {/eq}-component of the electric field produced by by the line of charge at point {eq}P {/eq} which is located at {eq}(x,y)= (a,0) {/eq}, where a {eq}6 cm {/eq}?

Electric Field from Charge Distributions:

The electric field due to a point charge q can be obtained using Coulomb's Law: {eq}\vec E = \dfrac {kq}{r^2} \hat r {/eq}, where k is the Coulomb's constant and r is the distance to the point of observation. The electric field due to a system of charges can be found using superposition principle: the total electric field is the sum of individual electric fields due to each point charge in the system.

Answer and Explanation:

Due to symmetry, the only component of electric field at the point P is the x-component; For this component, we can write:

{eq}E_x (P) = \displaystyle \int \limits_{-\infty}^{\infty} \dfrac {k\lambda_1 dy}{y^2 + a^2} \cdot \cos \theta = \int \limits_{-\infty}^{\infty} \dfrac {k\lambda_1 dy}{y^2 + a^2} \cdot \dfrac {a}{\sqrt{y^2 + a^2}} {/eq}

Making the substitution {eq}y = a \sinh x {/eq}, we obtain the following expression:

{eq}E_x (P) = \displaystyle \dfrac {k\lambda_1}{a} \int \limits_{-\infty}^{\infty} \dfrac {dx}{\cosh^2 x} {/eq}

Taking the integral, we obtain:

{eq}E_x (P) = \displaystyle \dfrac {k\lambda_1}{a} (\tanh x) \large|_{-\infty}^{\infty} {/eq}

Using the definition of hyperbolic tangent, we finally obtain:

{eq}E_x (P) = \displaystyle \dfrac {2k\lambda_1}{a} {/eq}

Calculation yields:

{eq}E_x (P) = \displaystyle \dfrac {2\cdot 9\times 10^9 \ N/(m^2 \cdot C^2) \cdot 4.6\times 10^{-4} \ C/ m}{0.06 \ m} = 1.38\times 10^8 \ N/C {/eq}


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from Physics 101: Intro to Physics

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