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Anna35 [415]
2 years ago
12

How many carbon atoms placed end to end would equal the width of a human hair

Physics
1 answer:
xz_007 [3.2K]2 years ago
3 0

The carbon atoms placed end to end would equal the width of a human hair are 1 million.

<h3>What is carbon?</h3>

The carbon is an element in the periodic table. It has atomic number 6 and the atomic mass 12 units.

Molecules anyway can shift in size from 0.1-0.5nm (nanometres) across so it would take a component of around 100,000 particles in width so that us might be able to see it with the unaided eye. Assuming that there's around 100,000 molecules in the width of one piece of hair, envision the number of there are in objects around you.

Learn more about carbon.

brainly.com/question/22530423

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3 years ago
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Two parallel plate capacitors 1 and 2 are identical except that capacitor 1 has charge +q on one plate and charge −q on the othe
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Answer:

a) the capacitance is the same for both capacitors.

b) The potential difference between the plates for the capacitor with charge +2q, is double of the one for the capacitor  with charge +q.

c) The electric field magnitude between the plates for the capacitor with charge +2q, is double of the one for the capacitor with charge +q

d) The energy stored between the plates for the capacitor with charge +2q, is 4 times the value for the one with charge +q  

Explanation:

a) The capacitance of a capacitor, by definition, is as follows:

C = \frac{q}{V}

Appying Gauss' Law to one of plates, it can be showed, that the capacitance (for a parallel plates capacitor) can be  expressed as follows:

C = ε*A / d

As it can be seen, it does not depend on the charge. so we conclude that the capacitance must be the same for both capacitors, due to they are identical except for the value of the charge on the plates.

b) By definition, as we said above, the capacitance is equal to the proportion between the charge of one of the plates, and the potential difference between them.

If this proportion must remain the same, and one of the capacitors has the double of  the charge than the other, the potential difference must be the double also.

c) Applying Gauss' law, to the surface of one of  the plates, and assuming a constant surface charge density σ, it can be  showed that the  electric field can be calculated as follows:

E*A = Q/ε₀ as σ=Q/A

⇒ E = σ/ε₀

As σ is directly proportional to the charge (being the area A the same), we conclude that the electric field for the capacitor with charge +2q must be the double than the one for the capacitior with charge +q.

c) The electric potential energy, stored between plates of a capacitor, can be written as follows:

Ue = \frac{1}{2} *\frac{q^{2}}{C}

If the capacitance remains the same, we can conclude that the electric potential energy for the capacitor with charge +2q, as the charge is raised to the 2nd power, must be 4 times the one for the capacitor with charge +q.

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