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ycow [4]
3 years ago
14

Three conducting plates, each of area A, are connected as shown.

Physics
1 answer:
Shkiper50 [21]3 years ago
7 0
You have effectively got two capacitors in parallel. The effective capacitance is just the sum of the two. 
Cequiv = ε₀A/d₁ + ε₀A/d₂ Take these over a common denominator (d₁d₂) 
Cequiv = ε₀d₂A + ε₀d₁A / (d₁d₂) Cequiv = ε₀A( (d₁ + d₂) / (d₁d₂) ) 
B) It's tempting to just wave your arms and say that when d₁ or d₂ tends to zero C -> ∞, so the minimum will occur in the middle, where d₁ = d₂ 
But I suppose we ought to kick that idea around a bit. 
(d₁ + d₂) is effectively a constant. It's the distance between the two outer plates. Call it D. 
C = ε₀AD / d₁d₂ We can also say: d₂ = D - d₁ C = ε₀AD / d₁(D - d₁) C = ε₀AD / d₁D - d₁² 
Differentiate with respect to d₁ 
dC/dd₁ = -ε₀AD(D - 2d₁) / (d₁D - d₁²)² {d2C/dd₁² is positive so it will give us a minimum} For max or min equate to zero. 
-ε₀AD(D - 2d₁) / (d₁D - d₁²)² = 0 -ε₀AD(D - 2d₁) = 0 ε₀, A, and D are all non-zero, so (D - 2d₁) = 0 d₁ = ½D 
In other words when the middle plate is halfway between the two outer plates, (quelle surprise) so that 
d₁ = d₂ = ½D so 
Cmin = ε₀AD / (½D)² Cmin = 4ε₀A / D Cmin = 4ε₀A / (d₁ + d₂)
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Due to the wave nature of light, light shined on a single slit will produce a diffraction pattern? Green light (520 nm) is shine
TiliK225 [7]

Answer:

Yes, it will produce a diffraction pattern.

a. 3.9 mm b. 1.95 mm

Explanation:

The light shined from a single slit will produce a diffraction pattern because,  the wavefront act as wavelets which generates its own wave according to Huygens principle. This therefore causes the diffraction pattern.

Given

wavelength of green light, λ = 520 nm = 520 × 10⁻⁹ m = 5.20 × 10⁻⁷ m

width of slit, d = 0.440 mm = 0.44 × 10⁻³ m = 4.4 × 10⁻⁴ m

Distance of slit from central maximum , D = 1.65 m

Distance of first minimum from central maximum, y = ?

a. The relationship between the slit width and wavelength is given by [tex} dsinθ = mλ [/tex]where d = slit width, θ = angular distance from central maximum, λ = wavelength of light and m = ±1, ±2, ±3...

The relationship between y and D is given by tanθ = y/D

Since θ is small, sinθ ≈ θ ≈ tanθ

so, dθ = mλ ⇒ θ = mλ/d = y/D

Therefore, y = mλD/d

Now, for the first minimum above the slit, m = +1 and for the first minimum below the slit, m = -1. So, y₁ =  λD/d and y₋₁ =  -λD/d. So, the width of the central maximum Δy is the difference between the first minima below and above the central maximum. So, Δy = y₁ - y₋₁ = λD/d -(-λD/d) = 2λD/d

Substituting the values from above, Δy= 2 × 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴ =  3900 × 10⁻⁶ m = 3.9 × 10⁻³ m = 3.9 mm

b. The first order fringe is the fringe located between the first minimum and the second minimum. From dsinθ = mλ and tanθ = y/D when θ is small, sinθ ≈ θ ≈ tanθ. So, y = mλD/d. Let m= 1 and m=2 be the first and second minima respectively. So,y₁ =  λD/d and y₂ =  2λD/d. The difference Δy₁ = y₂ - y₁ is the width of the first order fringe. Therefore, Δy₁ = 2λD/d - λD/d= λD/d. Substituting the values from above, we have

λD/d= 5.20 × 10⁻⁷ × 1.65/4.4 × 10⁻⁴= 1.95 × 10⁻³ m = 1.95 mm

7 0
3 years ago
An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 19 m lower vert
cupoosta [38]

Answer:

The compression in the spring is 5.88 meters.                

Explanation:

Given that,

Mass of the car, m = 39000 kg

Height of the car, h = 19 m

Spring constant of the spring, k=4.2\times 10^5\ N/m

We need to find the compression in the spring in stopping the ore car. It can be done by balancing loss in gravitational potential energy and the increase in elastic energy. So,

mgh=\dfrac{1}{2}kx^2

x is the compression in spring

x=\sqrt{\dfrac{2mgh}{k}} \\\\x=\sqrt{\dfrac{2\times 39000\times 19\times 9.8}{4.2\times 10^5}} \\\\x=5.88\ m

So, the compression in the spring is 5.88 meters.                                                                                                                  

6 0
3 years ago
radio waves travel at the speed of light, 3 x 10^8 m/s. the wavelength of a radio wave record is 200,000,000 hz is?
ludmilkaskok [199]

Answer:

1.5m

Explanation:

Speed of waves is given as the product of the wavelength and frequency. Sometimes when frequency is not given but the period is given, we get the frequency as the reciprocal of the period. The speed of waves is given in m/s, wavelength in m while frequency in Hz.

Speed, s= fw and making w the subject of formula,

w=\frac {s}{f}

Substituting 300, 000, 000 m/s for s and 200, 000, 000 for f then we obtain that

w=\frac {300000000}{200000000}=1.5 m

6 0
3 years ago
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To solve this problem it is necessary to apply the concepts related to the change of Energy in photons and the conservation of energy.

From the theory we could consider that the energy change is subject to

\Delta E = E_0 -W_f

Where

E_0 =Initial Energy

W_f = Energy loses

Replacing we have that

\Delta E = 1.6-0.8

\Delta E = 0.8eV

Therefore the Kinetic energy of the electron once it has broken free of the metal surface is 0.8eV

7 0
3 years ago
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Answer:

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5 0
3 years ago
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