An infinitely long line of charge with uniform density, rho???????? lies in y-z plane parallel to the zaxis at y=1m. (a) Find the potential

Question

An infinitely long line of charge with uniform density, rho???????? lies in y-z plane parallel to the zaxis at y=1m. (a) Find the potential VAB at point A (4m, 2m, 4m) in Cartesian coordinates with respect to point B (0,0,0). (b) Find E filed at point B.

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Latifah 6 months 2021-08-20T17:58:17+00:00 1 Answers 1 views 0

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    2021-08-20T18:00:03+00:00

    Answer with Explanation:

    We are given that

    Density=\rho l

    A(4m,2m,4m) and B(0,0,0)

    y=1 m

    a. Linear charge density=\frac{\rho l}{l}=\rho C/m

    Let a point P (0,1,4) on the line of charge  and point Q (0,1,0)

    Therefore,

    Distance AP=\sqrt{(4-0)^2+(2-1)^2+(4-4)^2}=\sqrt{17}

    Distance,BQ=\sqrt{(0-0)^2+(1-0)^2+(0-0)^2}=1

    Electric field for infinitely long line

    E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{r}

    Therefore, potential

    V_{BA}=-\int_{a}^{b}E\cdot dl

    V_{BA}=-\int_{\sqrt{17}}^{1}\frac{\rho}{2\pi \epsilon_0 r}\hat{r}\cdot \hat{r} dr

    V_{BA}=-\int_{\sqrt{17}}^{1}\frac{\rho}{2\pi \epsilon_0 r}dr

    V_{BA}=-\frac{\rho}{2\pi \epsilon_0}[\ln r]^{1}_{\sqrt{17}}

    V_{BA}=-\frac{\rho}{2\pi \epsilon_0}(ln 1-ln(\sqrt{17})=\frac{\rho}{2\pi \epsilon_0}(ln(\sqrt{17})

    V_{BA}=V_B-V_A

    V_{AB}=V_A-V_B=-V_{BA}=-\frac{\rho}{2\pi \epsilon_0}(ln(\sqrt{17})

    b.Electric field at point B

    E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{r}

    Unit vector r=-\hat{j}

    Therefore,

    E=\frac{\rho}{2\pi \epsilon_0 r}\cdot \hat{-j}

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