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cupoosta [38]
3 years ago
14

If you put a total of 8.05×106×106 electrons on an intially electrically neutral wire of length 1.03 m, what is the magnitude of

the electric field a perpendicular distance of 0.201 m away from the center of the wire?
Physics
1 answer:
olga_2 [115]3 years ago
6 0

Answer:

The magnitude of the electric field is 0.1108 N/C

Explanation:

Given;

number of electrons, e = 8.05 x 10⁶

length of the wire, L = 1.03 m

distance of the field from the center of the wire, r = 0.201 m

Charge of the electron;

Q = (1.602 x 10⁻¹⁹ C/e) x (8.05 x 10⁶ e)

Q = 1.2896 x 10⁻¹² C

Linear charge density;

λ = Q / L

λ = (1.2896 x 10⁻¹² C) / (1.03 m)

λ = 1.252 x 10⁻¹² C/m

The magnitude of electric field at r = 0.201 m;

E = (\frac{1}{4 \pi \epsilon_o} )\frac{ 2 \lambda}{r} \\\\E = k \frac{ 2 \lambda}{r}\\\\E = (8.89*10^9)*\frac{2*1.252*10^{-12}}{0.201}  \\\\E = 0.1108 \ N/C

Therefore, the magnitude of the electric field is 0.1108 N/C

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Answer:243joules

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Two different sources of radiation give the same dose equivalent in Sv. Does this mean that the radiation from each source has t
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5 0
2 years ago
The position of a particle is given by ~r(t) = (3.0 t2 ˆi + 5.0 ˆj j 6.0 t kˆ) m
Julli [10]

Answer:

v=(6ti+6k)\ m/s

Explanation:

Given that,

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r(t) = (3.0 t^2 i + 5.0j+ 6.0 tk) m

Let us assume we need to find its velocity.

We know that,

v=\dfrac{dr}{dt}\\\\=\dfrac{d}{dt}(3.0 t^2 i + 5.0j+ 6.0 tk) \\\\=(6ti+6k)\ m/s

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3 years ago
You walk with a velocity of 2 m/s north. You see a man approaching you, and from your frame of
solong [7]

Answer:

The velocity of the man from the frame of  reference of a stationary observer is, V₂ = 5 m/s

Explanation:

Given,

Your velocity, V₁ = 2 m/

The velocity of the person, V₂ =?

The velocity of the person relative to you, V₂₁ = 3 m/s

According to the relative velocity of two

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∴                               V₂ =  V₂₁ + V₁

On substitution

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Hence, the velocity of the man from the frame of reference of a stationary observe is, V₂ = 5 m/s

8 0
3 years ago
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