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 2.15 mm extends directly upward by 0.831 km and is coated with a 0.519 mm layer of water having resistivity 183 Ω m. If the potential difference between the two ends of the string is 166 MV, what is the current through the water layer?

Question

Palmer

Physics

8 July, 16:28

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 2.15 mm extends directly upward by 0.831 km and is coated with a 0.519 mm layer of water having resistivity 183 Ω m. If the potential difference between the two ends of the string is 166 MV, what is the current through the water layer?

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Answers (1)

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Angel Henderson · Answer 1

The current on the water layer = 1.64*10^-3A

Explanation:

Let's assume that the radius given for the string originates from the centre of the string. The equation for determining the current in the water layer is given by:

I = V * pi[ (Rwater + Rstring) ^2 - (Rstring) ^2 / (Resitivity * L)

I = [ 166*10^6 * 3.142[ (0.519*10^-4) + (2.15*10^-3]) ^2 - (2.15*10^-3) ^2] / (183 * 831)

I = [ 521572000 (4.848*10^6) - 4.623*10^-6] / 154566

I = 252.83 - (4.623*10^-6) / 154566

I = 252.83/154566

I = 1.64 * 10^-3A