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Bess [88]
3 years ago
13

3. A -4.00-uC charge lies 20.0cm to the right of a 2.00-uC charge on the x axis. What is the force on the 2.00-uC charge?

Physics
1 answer:
Sladkaya [172]3 years ago
3 0

<u>Answer:</u>

The force on the 2.00-uC charge is 36 \times 10^{10} \mathrm{N}

<u>Explanation:</u>

We know that force between two charges is given by Coulomb’s law,

\mathrm{F}=\mathrm{k} \frac{q 1 q 2}{r * r}

Where k = Coulomb’s constant =

9.0 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}

And q1 and q2 are the charges given to be = -4.00-uC and 2.00-uC charges

And r = distance between the charges = 20 cm = 0.2 m  

Substituting the given values in the formula we get force applied on 2.00\ \mu C charge,

F =  36 \times 10^{10} \mathrm{N} attractive force which is the required answer.

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Explanation:

Given that,

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A = 7.85 × 10^-3 m²

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The magnetic field is a function of time

B = 0.50exp(-20t) T

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We need to find the induced emf

This induced voltage, ε can be quantified by:

ε = −NdΦ/dt

Φ = BAcosθ, but θ = 90°, they are perpendicular

So, Φ = BA

ε = −NdΦ/dt = −N d(BA) / dt

A is a constant

ε = −NA dB/dt

Then, B = 0.50exp(-20t)

So, dB/dt = 0.5 × -20 exp(-20t)

dB/dt = -10exp(-20t)

So,

ε = −NA dB/dt

ε = −NA × -10exp(-20t)

ε = 10 × NA exp(-20t)

Now from ohms law, ε = iR

So, I = ε / R

I = 10 × NA exp(-20t) / R

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A hydrogen atom in the n=7 state decays to the n=4 state. what is the wavelength of the photon that the hydrogen atom emits? use
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A hydrogen atom in the n=7 state decays to the n=4 state. The wavelength of the photon that the hydrogen atom emits is 4592.59nm.

The Energy of photon is the energy possessed by a photon when it moves from a high energy level to a low energy level. It emits a photon of a certain wavelength. The following relation can be used to find out the relation between the energy levels and the energy possessed:

E = 13.6 × Z² (1/n₂² - 1/n₁²) eV

where, n₁ is the initial energy level i.e. n₁ =7

            n₂ is the higher energy level i.e. n₂ = 4

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Subsituting in above equation,

E = 13.6 (1/16 - 1/49) eV

E = 0.27 eV

We know that,

E = hc / λ  

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            λ is wavelength

On subsituting,

0.27 eV = 1240/ λ

⇒ λ = 4592.59 nm

Hence, the wavelength of photon emitted by Hydrogen atom is 4592.59nm.

Learn more about Energy of Photon here, brainly.com/question/2393994

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