Kepler's second law of planetary motion<span> describes the speed of a </span>planet<span> traveling in an elliptical orbit around the sun. It states that a line between the sun and the </span>planetsweeps equal areas in equal times. Thus, the speed of theplanet<span> increases as it nears the sun and decreases as it recedes from the sun.</span>
Answer:
11.962337 × 10^-4 N
Explanation:
Given the following :
Length L = 11.8
Charge = 29nC = 29 × 10^-9 C
Linear charge density λ = 1.4 × 10^-7 C/m
Radius (r) = 2cm = 2/100 = 0.02 m
Using the relation:
E = 2kλ/r ; F =qE
F = 2kλq/L × ∫dr/r
F = 2*k*q*λ/L × (In(0.02 + L) - In(0.02))
2*k*q*λ/L = [2 × (9 * 10^9) * (29 * 10^9) * (1.4 * 10^-7)]/ 0.118] = 6193.2203 × 10^(9 - 9 - 7) = 6193.2203 × 10^-7 = 6.1932203 × 10^-4
In(0.02 + 0.118) - In(0.02) = In(0.138) - In(0.02) = 1.9315214
Hence,
(6.1932203 × 10^-4) × 1.9315214 = 11.962337 × 10^-4 N
Explanation:
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