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ioda
3 years ago
13

17. (a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air ( 3.0×106

V/m ) if the plates are separated by 2.00 mm and 5.0×103 V a potential difference of is applied? (b) How close together can the plates be with this applied voltage?
Physics
1 answer:
sleet_krkn [62]3 years ago
5 0

Answer:

Explanation:

Distance between plates d = 2 x 10⁻³m

Potential diff applied = 5 x 10³ V

Electric field = Potential diff applied /  d

= 5 x 10³  / 2 x 10⁻³

= 2.5 x 10⁶ V/m

This is less than  breakdown strength for air  3.0×10⁶ V/m

b ) Let the plates be at a separation of d .so

5 x 10³ / d = 3.0×10⁶ ( break down voltage )

d = 5 x 10³  / 3.0×10⁶

= 1.67 x 10⁻³ m

= 1.67 mm.

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An amateur astronomer looks at the moon through a telescope with a 15-cm-diameter objective. What is the minimum separation betw
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Answer:

  y = 128.0 km

Explanation:

The minimum separation of two objects is determined by Rayleygh's diffraction criterion, which establishes that two bodies are solved if the first minino of diffraction of one coincides with the central maximum of the second, with this criterion the diffraction equation remains

                       

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In the case of circular openings, the equation must be solved in polar coordinates, leaving the expression, we use the approximation that the sine of tea is very small.

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to find the distance we can use trigonometry

             tan θ = y / L

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substituting

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let's calculate

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Let's reduce to km

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             y = 128.0 km

the correct answer is 120 km away

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3 years ago
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