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26 September, 00:01

The legend that Benjamin Franklin flew a kite as a storm approached is only a legend-he was neither stupid nor suicidal. Suppose a kite string of radius 2.00 mm extends directly upward by 0.660 km and is coated with a 0.500 mm layer of water having resistivity 150 Ω · m. If the potential difference between the two ends of the string is 130 MV, what is the current through the water layer

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  1. 26 September, 03:44
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    Answer: 0.93 mA

    Explanation:

    In order to calculate the current passing through the water layer, as we have the potential difference between the ends of the string as a given, assuming that we can apply Ohm's law, we need to calculate the resistance of the water layer.

    We can express the resistance as follows:

    R = ρ. L/A

    In order to calculate the area A, we can assume that the string is a cylinder with a circular cross-section, so the Area of the water layer can be written as follows:

    A = π (r22 - r12) = π ((0.0025) 2 - (0.002) 2) m2 = 7.07. 10-6 m2

    Replacing by the values, we get R as follows:

    R = 1.4 1010 Ω

    Applying Ohm's Law, and solving for the current I:

    I = V/R = 130 106 V / 1.4 1010 Ω = 0.93 mA
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