Question

1. A thin, long, straight filament, electrically charged with a constant linear charge density ?, the shortest distance from the observation point to the filament is p and the vacuum permittivity is ?0, the electric field created by this filament will be (3.5 Points) \frac{\lambda}{4\pi\epsilon_0\rho} \frac{\lambda}{2\pi\epsilon_0\rho} none \frac{3\lambda}{2\pi\epsilon_0\rho}

          1. A thin, long, straight filament, electrically
charged with a constant linear charge
density ?, the shortest distance from the
observation point to the filament is p and
the vacuum permittivity is ?0, the electric
field created by this filament will be
(3.5 Points)
\frac{\lambda}{4\pi\epsilon_0\rho}
\frac{\lambda}{2\pi\epsilon_0\rho}
none
\frac{3\lambda}{2\pi\epsilon_0\rho}

1. A thin, long, straight filament, electrically
charged with a constant linear charge
density ?, the shortest distance from the
observation point to the filament is p and
the vacuum permittivity is ?0, the electric
field created by this filament will be
(3.5 Points)
(λ)/(4πϵ0ρ)
(λ)/(2πϵ0ρ)
none
(3λ)/(2πϵ0ρ)

Added by Adam D.

University Physics with Modern Physics

Hugh D. Young 14th Edition

Instant Answer

Solved by Expert Hunza Gilgit

Step 1

The charge dq of this small section can be expressed as dq = Adl, where A is the linear charge density. The electric field dE created by this small section can be calculated using Coulomb's Law: dE = (1/4πε₀) * (dq/r²), where ε₀ is the vacuum Show more…

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A thin, long, straight filament, electrically charged with a constant linear charge density A. The shortest distance from the observation point to the filament is p, and the vacuum permittivity is Îµâ‚€. The electric field created by this filament will be: a) 4Ï€Îµâ‚€pA b) 2Ï€Îµâ‚€pA c) none d) 3Ï€Îµâ‚€pA