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A very long straight wire has charge per unit length 1.56×10−10 C/m .

At what distance from the wire is the magnitude of the electric field equal to 2.40 N/C ?
Use 8.85×10−12 C2/(N⋅m2) for the permittivity of free space, and use π=3.14159.

Respuesta :

akindeleot

Answer:

The magnitude of the electric field equal to 2.40 N/C at 1.1537m from the wire.

Explanation:

using Guass law,

(guessing that a cylinder of radius r and length L around wire such that wire is at centre )

E. A = qin / e0

E ( 2πr L ) = (1.56 x [tex]10^{-10}[/tex] x L) / (8.85 x [tex]10^{-12}[/tex])

E = (1.56 x [tex]10^{-10}[/tex] ) / (2πr x 8.85 x  [tex]10^{-12}[/tex])

so 2.40 = (1.54 x [tex]10^{-10}[/tex] ) / (2πr x 8.85 x [tex]10^{-12}[/tex])

2.40 (6.284r) = 0.174 x 10²

15.0816r = 17.4

r = 1.1537m

Cricetus

The distance will be "1.1537 m".

Given:

  • Electric field = [tex]2.40 \ N/C[/tex]
  • Charge = [tex]1.56\times 10^{-10} \ C/m[/tex]

By using the Gauss law, we get

→ [tex]E.A = \frac{q_{in}}{e_0}[/tex]

or,

→  [tex]E(2 \pi r L) = \frac{1.56\times 10^{-10}\times L}{8.85\times 10^{-12}}[/tex]

               [tex]E = \frac{1.56\times 10^{-10}}{2 \pi r\times 8.85\times 10^{-12}}[/tex]

            [tex]2.40 = \frac{1.54\times 10^{-10}}{2 \pi r\times 8.85\times 10^{-12}}[/tex]

[tex]2.40\times 6.284 r= 0.174\times 10^2[/tex]

      [tex]15.0816 r = 17.4[/tex]

                 [tex]r = \frac{17.4}{15.0816 }[/tex]

                    [tex]= 1.1537 \ m[/tex]

Thus the above approach is right.  

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