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vitfil [10]
4 years ago
6

The electric potential, when measured at a point equidistant from two particles that have charges equal in magnitude but of oppo

site sign, is
Physics
1 answer:
olga2289 [7]4 years ago
4 0

Answer:

V = 0

Explanation:

To find the potential at the middle of the two charges with opposite signs, you use the following formula:

V_T=V_1+V_2=k\frac{q}{r}-k\frac{q}{r}=0V      (1)

where you have used the fact that the charges are the same and the distances are the same.

The electric potential is zero at the point in the middle of the two charges

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Lead atoms occupy a volume of 3 x 10-29 m3. Each atom contributes two free electrons. Calculate the Fermi velocity of lead.
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<u>Answer:</u> The Fermi velocity of lead is 64.4 km/s.

<u>Explanation:</u>

To calculate the Fermi velocity, we use the equation:

V_f=\frac{h}{2\pi m_e}(\frac{3\pi^2N}{V})^{1/3}

where,

h = Planck's constant = 6.62\times 10^{-34}Js

m_e = mass of electron = 9.1\times 10^{-31}kg

N = Number of atoms present in per volume of atom multiplied by number of electrons present in given atom = \frac{2\times N_A\times V}{M}

N_A = Avogadro's number = 6.022\times 10^{26}mol^{-1}    (When the mass is in kilograms)

V = Volume = 3\times 10^{-29}m^3

M = molecular weight of lead = 207.2 g/mol

Putting values in above equation, we get:

V_f=\frac{6.62\times 10^{-34}}{2\times 3.14\times (9.1\times 10^{-31})}(\frac{3\times (3.14)^2\times (2\times 6.022\times 10^{23}\times 3\times 10^{-29})}{3\times 10^{-29}\times 207.2})^{1/3}

V_f=0.0644\times 10^6m/s=64.4km/s     (Conversion factor: 1 km = 1000 m)

Hence, the Fermi velocity of lead is 64.4 km/s

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