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

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

answer:

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