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frozen [14]
3 years ago
7

The electric field strength E 0 is measured at a perpendicular distance R from an infinitely large, thin sheet that contains a u

niform positive charge density σ . In terms of the stated properties and the permittivity of free space ε 0 , write an expression for the electric field strength E at a perpendicular distance 2 R from the sheet.
Physics
1 answer:
Brilliant_brown [7]3 years ago
7 0

Answer:

Electric Field Strength E₀ = E₀ = Constant

Explanation:

The Electric Filed Strength E₀ to an Infinite uniformly charge large sheet is constant how far is it i.e. it is independent of the distance away from the uniformly charge sheet.

Formula: E₀ = σ / 2 ε₀  

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a. What proportion of resistors have resistances less than 90 Ω? b. Find the mean resistance. c. Find the standard deviation of
Anettt [7]

Answer:

a) 0.0625 = 6.25%

b) 106.67 Ω

c) 9.43 Ω

d) 1

Explanation:

The probability distribution is given as

f(x) = (x - 80)/800 for 80 < x < 120

f(x) = 0 otherwise.

f(x) = (x/800) - (0.1)

a) Proportion of resistors with resistance less than 90 Ω

P(X < 90) = ∫⁹⁰₈₀ f(x) dx

∫⁹⁰₈₀ f(x) dx = ∫⁹⁰₈₀ [(x/800) - (0.1)]

= [(x²/1600) - 0.1x]⁹⁰₈₀

= [(90²/1600) - 0.1(90)] - [(80²/1600) - 0.1(80)]

= (5.0625 - 9) - [4 - 8]

= -3.9375 + 4 = 0.0625 = 6.25%

b) The mean is given by the expected value expression E(X) = = Σ xᵢpᵢ (with the sum done all over the data set for each variable and its corresponding probability)

It can be written in integral form as

Mean = ∫¹²⁰₈₀ xf(x) dx (with the integral done all over the probability function, i.e. from, 80 to 120)

Mean = ∫¹²⁰₈₀ x[(x/800) - (0.1)] dx

= ∫¹²⁰₈₀ [(x²/800) - (0.1x)] dx

= [(x³/2400) - (0.05x²)]¹²⁰₈₀

= [(120³/2400) - (0.05(120²)] - [(80³/2400) - (0.05(80²)]

= [720 - 720] - [213.33 - 320] = 106.67 Ω

c) Standard deviation = √(variance)

Variance = Var(X) = Σx²p − μ²

μ = mean = expected value = 106.67 Ω

Σx²p = ∫¹²⁰₈₀ x²f(x) dx = ∫¹²⁰₈₀ x² [(x/800) - (0.1)] dx = ∫¹²⁰₈₀ [(x³/800) - (0.1x²)] dx

= [(x⁴/3200) - (0.0333x³)]¹²⁰₈₀

= [(120⁴/3200) - (0.0333(120³)] - [(80⁴/3200) - (0.0333(80)³)]

= (64800 - 57600) - (12800 - 17066.667)

= 11466.667

Variance = 11466.667 - 106.67² = 88.85

Standard deviation = √88.85 = 9.43 Ω

d) Cdf = sum of probabilities over the entire probability function

Cdf = ∫¹²⁰₈₀ f(x) dx = ∫¹²⁰₈₀ [(x/800) - (0.1)] dx

= [(x²/1600) - 0.1x]¹²⁰₈₀ = [(120²/1600) - 0.1(120)] - [(80²/1600) - 0.1(80)] = (9 - 12) - (4 - 8) = -3+4 = 1 as it should be!!!

Hope this Helps!!!

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Lapatulllka [165]

Answer:

Therefore the correct statement is B.

Explanation:

In the interference and diffraction phenomena, the natural wave of electromagnetic radiation must be taken into account, the wave front that advances towards the slit can be considered as when it reaches it behaves like a series of wave emitters, each slightly out of phase from the previous one, following the Huygens principle that states that each point is compiled as a source of secondary waves.

The sum of all these waves results in the diffraction curve of the slit that has the shape

      I = Io sin² θ /θ²

Where the angle is a function of the wavelength and the width of the slit.

From the above, the interference phenomenon can be treated as the sum of two diffraction phenomena displaced a distance equal to the separation of the slits (d)

Therefore the correct statement is B

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