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pav-90 [236]
3 years ago
15

A very long wire carries a uniform linear charge density of 5 nC/m. What is the electric field strength 13 m from the center of

the wire at a point on the wire's perpendicular bisector? (ε 0 = 8.85 × 10-12 C2/N · m2)
Physics
1 answer:
kipiarov [429]3 years ago
5 0

Answer:

E=6.91 N/C

Explanation:

Given that

Linear Charge density ,λ = 5 nC/m

Distance ,R= 13 m

We know that formula for long wire to find electric field

E=\dfrac{\lambda }{2\pi \varepsilon _0R}

E=Electric field

R=Distance

εo=8.85 x 10⁻¹² C²/N.m²

λ=Linear Charge density

Now by putting the values

E=\dfrac{5\times 10^{-9}}{{2\times \pi \times 8.85\times 10^{-12}\times 13}}

E=6.91 N/C

Therefore the electric filed at distance 13 m will be 6.91 N/C

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The distance traveled by the particle at the given time interval is 0.28 m.

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Learn more about distance here: brainly.com/question/23848540

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