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dlinn [17]
3 years ago
11

Air "breaks down" when the electric field strength reaches 3 × 106 n/c, causing a spark. A parallel-plate capacitor is made from

two 5.5 cm × 5.5 cm electrodes.
Physics
2 answers:
Lera25 [3.4K]3 years ago
3 0

Electric field between the plates of parallel plate capacitor is given as

E = \frac{Q}{A\epsilon_0}

here area of plates of capacitor is given as

A = 0.055 * 0.055

A = 3.025 * 10^{-3}

also the maximum field strength is given as

E = 3 * 10^6 N/C

now we will plug in all data to find the maximum possible charge on capacitor plates

3 * 10^6 = \frac{Q}{3.025 * 10^{-3}*8.85 * 10^{-12}}

Q = 8.03 * 10^{-8} C

so the maximum charge that plate will hold will be given by above

victus00 [196]3 years ago
3 0
<h2>Complete Question:</h2>

Air "breaks down" when the electric field strength reaches 3×10⁶N/C, causing a spark. A parallel-plate capacitor is made from two 5.5cm × 5.5cm plates.  

How many electrons must be transferred from one disk to the other to create a spark between the disks?

<h2>Answer:</h2>

5.02 x 10¹¹

<h2>Explanation:</h2>

The electric field E of a parallel-plate capacitor is given by;

E = Q / (A ε₀)        --------------------------(i)

Where;

Q = the charge on the plates

A = Area of any of the plates

ε₀ = electric constant =  8.85 x 10⁻¹²F/m

<em>From the question;</em>

E = 3 x 10⁶N/C

A = 5.5cm x 5.5cm = 30.25cm² = 0.003025m²

<em>Substitute these values into equation (i) as follows;</em>

3 x 10⁶ = Q / (0.003025 x 8.85 x 10⁻¹²)

<em>Solve for Q;</em>

Q = 3 x 10⁶ x 0.003025 x 8.85 x 10⁻¹²

Q = 8.03 x 10⁻⁸

The charge on the capacitor is therefore, 8.03 x 10⁻⁸C

Now, to get the number of electrons that must be transferred from one disk to the other to create a spark between the disks, we use the relation;

Q = N x E               ----------------------(ii)

Where;

Q = the charge = 8.03 x 10⁻⁸C                            [as calculated above]

N = the number of electrons

E = the charge of one electron = 1.6 x 10⁻¹⁹C    [a known value]

<em>Substitute these values into equation (ii) as follows;</em>

8.03 x 10⁻⁸ = N x 1.6 x 10⁻¹⁹

<em>Solve for N;</em>

N = [8.03 x 10⁻⁸] / [1.6 x 10⁻¹⁹]

N = 5.02 x 10¹¹

Therefore, the number of electrons that must be transferred from one disk to the other to create a spark between the disks is 5.02 x 10¹³

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Answer:

The temperature beyond which the substance overflows the container is 86.23°C.

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Professor Hosney took the second grade students to the laboratory to perform an experiment. He took a 1000ml capacity container at a temperature of 68oF and poured 980 ml of a substance at 20oC into it. While placing the set to heat, he consulted a table where he found the volumetric expansion coefficient of the substance, 4 x 10-4 ºC-1 and the linear expansion coefficient of the container material, 3 x 10-5 ºC-1. Hosney then asked students to determine the temperature from which the substance would overflow. A student then asked, what is the melting temperature of the substance, and the teacher answered promptly 290.8 K. What is the temperature from which the substance will overflow?

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where

γ = coefficient of volume expansion

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(ΔT) = change in temperature.

At the temperature where the substance will overflow, the volume of the substance and the container will both be the same.

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We will still leave ΔT as ΔT

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New volume of the substance at that temperature = V₀ + ΔV₁ = 980 + 0.392ΔT

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V₀ = Initial volume = 1000 mL

(ΔT) = change in temperature = (T - 20) (note that 68°F = 20°C)

We will still leave ΔT as ΔT

ΔV₂ = (9 × 10⁻⁵) × 1000 × ΔT

ΔV₂ = 0.09 ΔT

New volume of the container at that temperature = V₀ + ΔV₂ = 1000 + 0.09 ΔT

At the temperature where overflow occurs, the two volumes are initially first the same.

980 + 0.392ΔT = 1000 + 0.09 ΔT

0.392ΔT - 0.09ΔT = 1000 - 980

0.302ΔT = 20

ΔT = (20/0.302) = 66.23°C

T - 20° = 66.23°

T = 66.23 + 20 = 86.23°C

The temperature beyond which the substance overflows the container is 86.23°C.

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