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scoray [572]
3 years ago
15

What is a sound of electronic?​

Physics
1 answer:
uysha [10]3 years ago
6 0

Answer:

sound of electronic means

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Two electrons with a charge of magnitude 1.6×10-19 C in an atom are separated by 1.5×10-10 m, the typical size of an atom. What
vesna_86 [32]

Answer:

1.02\cdot 10^{-8} N, repulsive

Explanation:

The magnitude of the electric force between two charged particles is given by Coulomb's law:

F=k\frac{q_1 q_2}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q_1, q_2 are the two charges of the two particles

r is the separation between the two charges

The force is:

- repulsive if the two charges have  same  sign

- Attractive if the two charges have opposite signs

In this problem, we have two electrons, so:

q_1=q_2=1.6\cdot 10^{-19}C is the magnitude of the two electrons

r=1.5\cdot 10^{-10} m is their separation

Substituting into the formula, we find the electric force between them:

F=(8.99\cdot 10^9)\frac{(1.6\cdot 10^{-19})^2}{(1.5\cdot 10^{-10})^2}=1.02\cdot 10^{-8} N

And the force is repulsive, since the two electrons have same sign charge.

4 0
3 years ago
What do you mean by peltier coefficient? ​
devlian [24]

Answer:

The Peltier coefficient is a measure of the amount of heat carried by electrons or holes

Explanation:

4 0
4 years ago
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3 years ago
What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?
konstantin123 [22]

Answer:

I=2.71\times 10^{-5}\ A

Explanation:

A 6.0-cm-diameter parallel-plate capacitor has a 0.46 mm gap.  

What is the displacement current in the capacitor if the potential difference across the capacitor is increasing at 500,000V/s?

Let given is,

The diameter of a parallel plate capacitor is 6 cm or 0.06 m

Separation between plates, d = 0.046 mm

The potential difference across the capacitor is increasing at 500,000 V/s

We need to find the displacement current in the capacitor. Capacitance for parallel plate capacitor is given by :

C=\dfrac{A\epsilon_o}{d}\\\\C=\dfrac{\pi r^2\epsilon_o}{d}, r is radius

Let I is the displacement current. It is given by :

I=C\dfrac{dV}{dt}

Here, \dfrac{dV}{dt} is rate of increasing potential difference

So

I=\dfrac{\pi r^2\epsilon_o}{d}\times \dfrac{dV}{dt}\\\\I=\dfrac{\pi (0.03)^2\times 8.85\times 10^{-12}}{0.46\times 10^{-3}}\times 500000\\\\I=2.71\times 10^{-5}\ A

So, the value of displacement current is 2.71\times 10^{-5}\ A.

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You can, just tell them
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