Two parallel circular rings of radius R have their centres in the X axis separated by a distance L. If each ring carries a uniformly distrib

Two parallel circular rings of radius R have their centres in the X axis separated by a distance L. If each ring carries a uniformly distributed charge Q,find the electric field at points along the X axis

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  1. Answer:

    E” =  Q/4πε₀√[(x² + R²)]³(x – (L – x)/√[(L – 2x)L/(x² + R²) + 1]³})

    Explanation:

    The electric field due to a charged ring of radius R at a distance x from the center of the ring when the axis of the ring is located on the x – axis is

    E = Qx/4πε₀[√(x² + R²)]³

    Since the rings are separated by a distance L, the electric field at point x due to the second ring is E’ = -Q(L – x)/4πε₀[√((L – x)² + R²)]³. It is negative since it points in the negative x – direction.

    So, the resultant electric field at x is E” = E + E’ = Qx/4πε₀[√(x² + R²)]³ + {-Q(L – x)/4πε₀[√((L – x)² + R²)]³}

    E” =  Qx/4πε₀√[(x² + R²)]³ – Q(L – x)/4πε₀√[((L – x)² + R²)]³

    E” =  Q/4πε₀(x/√[(x² + R²)]³ – (L – x)/√[((L – x)² + R²)]³})

    E” =  Q/4πε₀(x/√[(x² + R²)]³ – (L – x)/√[(L² – 2Lx + x² + R²)]³})

    E” =  Q/4πε₀(x/√[(x² + R²)]³ – (L – x)/√[(L – 2x)L + (x² + R²)]³})

    E” =  Q/4πε₀√[(x² + R²)]³(x – (L – x)/√[(L – 2x)L/(x² + R²) + 1]³})

    So, the electric field at points along the x axis is

    E” =  Q/4πε₀√[(x² + R²)]³(x – {(L – x)/√[(L – 2x)L/(x² + R²) + 1]³})

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