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sladkih [1.3K]
4 years ago
5

A very long wire carries a uniform linear charge density of 7.0 nC/m. What is the electric field strength 16.0 m from the

Physics
1 answer:
jok3333 [9.3K]4 years ago
6 0

Electric field due to a long wire is given by

E = \frac{2 k \lambda}{r}

here

k =  9 * 10^9

\lambda = 7*10^{-9}

r = 16 m

E = \frac{2*9*10^9 *7*10^{-9}}{16}

E = 7.875 N/C

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A particle with charge q = +5e and mass m = 8.2×10-26 kg is injected horizontally with speed 1.1×106 m/s into the region between
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Answer:

The third charge must be placed 0.548 m from q₁.

Explanation:

Let r = 3m  be the distance between charge q₁ and q₂.

Let x be the distance between charge q₁ and charge q₃ (the third positive charge)

Then r - x is the distance between charge q₂ and q₃

Let the electrostatic force between q₁ and q₃ be F = kq₁q₃/x²

Let the electrostatic force between q₂ and q₃ be F' = kq₂q₃/(r - x)²

Since F + (-F') = 0 (the signs on the forces are different since the forces are in opposite directions)which is required when the net force on q₃ is zero, then

F - F' = 0

F = F'

So, kq₁q₃/x² = kq₂q₃/(r - x)²

q₁/x² = q₂/(r - x)²

[(r - x)/x]² = q₂/q₁

taking square-root of both sides,

(r - x)/x = ±√q₂/q₁

r/x - 1 = ±√q₂/q₁

r/x = 1 ±√q₂/q₁

x = r/(1 ±√q₂/q₁)

substituting the values of the variables r = 3 m, q₁ = 0.50 nC and q₂ = 10 nC

x = 3 m/(1 ±√10 nC/0.5 nC)

x = 3 m/(1 ±√20)

x = 3 m/(1 ± 4.472)

x = 3 m/5.472 or 3 m/-3.472

x = 0.548 m or -0.864 m

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Since we are concerned about the line of charge that connects q₁ and q₂, the third charge must be placed 0.548 m from q₁.

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