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Gelneren [198K]
3 years ago
12

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500

Physics
1 answer:
padilas [110]3 years ago
5 0

Answer:

I = 1.21x10^-5 A

Explanation:

You are missing the first part of the problem. This is an example, but it will give you the idea of how to solve yours with your data.

The first part is like this:

<em>A      4.0 cm  diameter parallel plate capacitor has a  0.44 m  m    gap. What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000 V/s?</em>

Now with this, we can solve the problem.

In order to do this, we need to use the following expression:

q = CV (1)

Where:

C: Capacitance of a parellel capacitor (in Faraday)

q: charge of plate or capacitor (In coulombs)

V: voltage in Volts.

However, we need is the current, and we have data of potential difference, so, all we have to do is divide the expression between time so:

q/t = CV/t

And the current is q/t, thus:

I = C * V/t (2)

And finally, Capacitance C with two plates of area A separated by a distance d is:

C = Eo*A/d (3)

Where:

Eo = constant equals to 8.85x10^-12 F/m.

A = Area of the plate, in this case, πr²

d = gap of the capacitor.

Let's calculate first the Capacitance using equation (3):

C = 8.85x10^-12 * π * (0.04/2)² / 0.00046 = 2.42x10^-11 F

Now, it's time to use equation (2) and solve for I:

I = 2.42x10^-11 * 500,000

I = 1.21x10^-5 A

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Imagine an alternate universe where the value of the Planck constant is . In that universe, which of the following objects would
HACTEHA [7]

Question: The planck constant was not given. In this calculation, planck constant of 6.62607*10^-9 Js  is used for the calculation.

Answer:

(a) A virus -------------Classical

(b) A buckyball -----Classical

(c) A mosquito ------ Quantum

(d) A turtle  ------------Quantum

Explanation:

 Calculating the wavelength using the formula;

λ= h/(mv)

where

λ= Wavelength

h = Planck Constant = 6.62607*10^-9 Js

m = mass in kg

v = velocity in m/s

Virus size = 280. nm = 2.80*10⁻⁷ m

a)

A Virus:

m = 9.4 x 10-17 g 9.4*10⁻²⁰ kg

v = 0.50 µm/s = 5 *10⁻⁷ m/s

h = 6.62607*10^-9 Js

Virus size = 280 nm = 2.80*10⁻⁷ m

Substituting into the formula; we have

λ= h/(mv)

λ= 6.62607*10^-9/ (9.4*10⁻²⁰* 5 *10⁻⁷)

  = 6.62607*10^-9/4.7*10^-26

  = 1.4*10^17 m

Classical : Wavelength is bigger than it's size

(b)

A buckyball

m = 1.2 x 10-21 g = 1.2 *10⁻²⁴ kg

V = 37 m/s

Size = 0.7 nm = 7*10⁻¹⁰ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ ( 1.2 *10⁻²⁴* 37)

  =  6.62607*10^-9/4.44*10^-23

  = 1.49 *10^14 m

Classical : Wavelength is bigger than it's size

(c)

A mosquito

Mass = 1.0 mg = 1*10⁻⁶ kg

v = 1.1 m/s

Size =  6.3 mm = 6.3*10⁻³ m

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  1*10⁻⁶* 1.1)

  =  6.62607*10^-9/1.1*10^-6

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Quantum Approach: The wavelength and the size are comparable

(d)

A turtle

Mass = 710. g = 0.71 kg

Size =  22. cm = 0.22 m

V =  2.8 cm/s. = 0.028 m/s

Substituting into the formula, we have

λ= h/(mv)

λ= 6.62607*10^-9/ (  0.71* 0.028)

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Quantum Approach: The wavelength and the size are comparable

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jeyben [28]

Answer:

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Explanation:

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v = Sound wave velocity = 330 m/s

v_o = Observer velocity = X m/s

v_s = Source velocity = 0 m/s (Assuming the source is stationary)

Which gives;

20 = \dfrac{330- v_o}{330+0} \times 21

330 - v_o = (20/21)*330

v_o = 330 - (20/21)*330 = 15.7 m/s

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