What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.

Answers 2

Answer:

 λ = -47 nC / m  

Explanation:

The missing question is as follows:

" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.  "

Given:

- The Diameter of the power line D = 2.2 cm

- The distance between two ends of power line L = 1.9m

- The potential difference across two ends V = 3.8 KV

Find:

What is the magnitude of the line charge density on the power line?

Solution:

- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:

                                      V = 2*k*λ*Ln( D / L )

Where,

k : Coulomb's Constant = 8.99*10^9

 λ : The line charge density

- Re-arrange and solve for λ:

                                     λ = V / 2*k*Ln( D / L )

Plug in the values:

                                    λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )

                                    λ = -4.74022*10^-8 C / m

                                   λ = -47 nC / m  

AI generated Answer

Answer: 3.3 µC/m Explanation: The magnitude of the line charge density on a power line is calculated by dividing the total charge on the line (Q) by its length (L). In this case, the charge on the line is 3.3 µC and the length of the line is 1 m. Therefore, the line charge density can be calculated as: Line charge density = Q/L = 3.3 µC / 1 m = 3.3 µC/m.
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