The electrostatic force between two charges Q1 and q is given by
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges
Re-arranging the formula, we have
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.
Planck find the correct curve for the specturm of light emitted by a hot object by vibrational energies of the atomic resonators were quantized.
<h3>Briefing :</h3>
- The energy density of a black body between λ and λ + dλ is the energy E=hc/λ of a mode times the density of states for photons, times the probability that the mode is occupied.
- This is Planck's renowned equation for a black body's energy density.
- According to this, electromagnetic radiation from heated bodies emits in discrete energy units or quanta, the size of which depends on a fundamental physical constant (Planck's constant). The basis of infrared imaging is the correlation between spectral emissivity, temperature, and radiant energy, which is made possible by Planck's equation.
Learn more about the Planck's constant with the help of the given link:
brainly.com/question/27389304
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Answer:
4.4345× 10^-7V
Explanation:
The computation of the half voltage for a 1.2T magnetic field applied is shown below
The volume of one mole of copper is
v = m ÷p
= 63.5 ÷ 8.92
= 7.12cm
Now the density of free electrons in copper is
n = Na ÷ V
= 6.02 × 10^23 ÷ 7.12
= 8.456× 10^28/m^3
Now the half voltage is
= IB ÷ nqt
= (5 × 1.20) ÷ (8.456× 10^28 × 1.6 × 10^-19 × 0.1× 10^-2)
= 4.4345× 10^-7V
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