Answer:
The current is [tex]I = 1.1434*10^{-5}}\ A[/tex]
Explanation:
From the question we are told that
The radius of the kite string is [tex]R = 2.02 mm = 0.00202 \ m[/tex]
The distance it extended upward is [tex]D = 0.823 km = 823 \ m[/tex]
The thickness of the water layer is [tex]d = 0.506 mm = 0.000506 \ m[/tex]
The resistivity is [tex]\rho = 159\ \Omega \cdot m[/tex]
The potential difference is [tex]V = 186 MV = 186 *10^{6} \ V[/tex]
Generally the cross sectional area of the water layer is mathematically represented as
[tex]A = \pi r^2[/tex]
Here r is mathematically represented as
[tex]r = [(R + d ) - R][/tex]
=> [tex]r = [(0.00202 + 0.000506 ) - 0.00202][/tex]
=> [tex]r = 0.000506[/tex]
=> [tex]A = 3.142 * [0.000506]^2 [/tex]
=> [tex]A = 8.0447*10^{-7}\ m^2 [/tex]
Generally the resistance of the water is mathematically represented as
[tex]R = \frac{\rho * D }{A}[/tex]
=> [tex]R = \frac{159 *823 }{8.0447*10^{-7}}[/tex]
=> [tex]R = 1.62662 * 10^{11} \ \Omega [/tex]
Generally the current is mathematically represented as
[tex]I = \frac{V}{R}[/tex]
=> [tex]I = \frac{186 *10^{6} }{1.62662 * 10^{11}}[/tex]
=> [tex]I = 1.1434*10^{-5}}\ A[/tex]
