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juin [17]
4 years ago
11

With a thunderstorm brewing, an electric field of magnitude 2.0 × 102 newtons/coulomb exists at a certain point in the earth’s a

tmosphere. If an electron at that point has a charge of -1.6 × 10-19 coulombs, what is the magnitude of the force on the electron as a result of the electric field?
A) 2.0 × 10^-6 newtons
B) 3.2 × 10^-17 newtons
C)5.8 × 10^5 newtons
D)6.4 × 10^-3 newtons
E) 1.7 × 10^4 newtons
Physics
1 answer:
hichkok12 [17]4 years ago
3 0
(2.0 x 10² N/Coul) x (-1.6 x 10⁻¹⁹ Coul) = -3.2 x 10⁻¹⁷ N

The magnitude is 3.2 x 10⁻¹⁷ Newton.
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A baseball is thrown a velocity of 20 m/s. It takes the 20 kg ball 10 seconds 5 points
liberstina [14]

Answer: 40 Newton

Explanation:

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Velocity of baseball = 20 m/s.

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7 0
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A sample of N-type silicon is at the room temperature. When an electric field with a strength of 1000V /cm is applied to the sam
V125BC [204]

Answer:

a) 222/cm^3

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Explanation:

(a) From the velocity and the applied electric field, we can calculate the mobility of holes:

υdp = µpε, µp = υdp/ε = 2×10^5/1000

= 200cm2/V.s

From a), we find Nd is equal to 4.5×1017/cm3

. Hence,

n = Nd = 4.5×10^17/cm3

, and p = ni^2/n

= ni^2/Nd

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.

Clearly, the minority carrier is hole.

(b) The Fermi level with respect to Ec is

Ef = Ec - kTln(Nd/Nc) = Ec - 0.107 eV.

(c) R = ρL/A. Using Equation, we first calculate the resistivity of the sample:

σ = q(µn n + µp p) ≈ qµn n = 1.6×10^-19 × 400 × 4.5×10^17 = 28.8/Ω-cm, and

ρ= σ

-1 = 0.035 Ω-cm.

Therefore, R = (0.035) × 20µm / (10µm× 1.5µm) = 467 Ω.

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