Anonymous
Anonymous asked in Science & MathematicsPhysics · 10 months ago

The potential difference between the surface of a 2.4 cm -diameter power line and a point 1.4 m distant is 4.8 kV?

The potential difference between the surface of a 2.4 cm -diameter power line and a point 1.4 m distant is 4.8 kV.

What is the magnitude of the line charge density on the power line? Answer in nC/m.

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  • NCS
    Lv 7
    10 months ago
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    I've spent some time researching this and I'm pretty sure this is right:

    V = ∫E dr = ∫ (λ / 2πrε₀) dr = (λ / 2πε₀)*ln(R/r) = 4800 V

    The only question I have is "what is R" -- in other words, what does "1.4 m distant" mean. Is it distance from the surface of the conductor, or distant from its center?

    If from the center, then

    λ = 4800V * 2π*8.85e-12 C²/N·m² / ln(1.4/0.012) = 5.6e-8 C/m

    which is 56 nC/m ◄

    and now I see the mistake I made before -- I multiplied by the ln term instead of divided.

    If "distant" means "from the surface of the power line," then

    λ = 4800V * 2π*8.85e-12 C²/N·m² / ln(1.412/0.012) = 5.6e-8 C/m

    so the same.

    Whew!

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