Question 1. Kiting during a storm . The legend that Benjamin Franklin flew a kit

Question

Question

1. Kiting during a storm. 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 2.07 mm extends directly upward by 0.817 km and is coated with a 0.517 mm layer of water having resistivity 171 ?·m. If the potential difference between the two ends of the string is 167 MV, what is the current through the water layer? The danger is not this current but the chance that the string draws a lightning strike, which can have a current as large as 500 000 A (way beyond just being lethal).

Question

2. In the figure a battery of potential difference V = 42 V is connected to a resistive strip of resistance R = 8.3 ?. When an electron moves through the strip from one end to the other, (a) in which direction in the figure does the electron move, (b) how much work is done on the electron by the electric field in the strip, and (c) how much energy is transferred to the thermal energy of the strip by the electron?

Explanation / Answer

ANSWER

Problem 1

Get the resistance of the "tube" of water using

   R = L/ A        

We need to determine A. This is the cross section area of the water minus the cross section area of the string.

The water "tube" has outside radius of   2.07 + 0.517   = 1.553 mm So,

   area of water = r2 = * 1.5532 = 7.58 mm2

   area of string = *2.072 = 13.46 mm2

   A = 13.46 - 7.58 = 5.88 mm2 = 5.88 x10-6 m2

And then

R = 171 * 817 / 5.88 x10-6     = 23759.69 x 106 ohms

and then  

     current = V / R = 167 x 106 / 23759.69 x106   =    0.007029 A = 7.029 mA   ANS.

Problem 2

a) Counter clockwise   ANS.

b)

Potential difference of the battery , V = 42 V

Resistance of resistive strip , R = 8.3

The charge of electron is, q = 1.6 *10^-19 C

Work done on the electron by the electric field is

         W = q (V) = (1.6 *10^-19 C) * (42V)

W = 67.2 *10^-19 J

In electron volts, 1 eV = 1.6 *10^-19 J

Thus, W = ( 67.2 * 10^-19 J) / (1.6 *10^-19 eV /J)

W = 42 eV   ANS.

Part c)

The energy transferred to the thermal energy of the strip is equal to the work done on the electron by the electric field,

       U = W = 42 eV   ANS.

Regards!!!

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