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Artemon [7]
3 years ago
13

Which of the following changes will increase the capacitance of a parallel-plate capacitor? (There could be more than one correc

t choice.)
A) increase the charge on the plates
B) decrease the poten5al between the plates
C) increase the potential between the plates
D) introduce a dielectric material between the plates
E) decrease the separation between the plates
Physics
1 answer:
Ber [7]3 years ago
7 0

As we know that capacitance of parallel plate capacitor is given as

C = \frac{\epsilon_0 A}{d}

here we know that the gap between the plates of capacitor is filled by air or vacuum

now if we fill the space between the plates by dielectric medium then we will have

C_{med} = \frac{k \epsilon_0 A}{d}

so it will increase the capacitance of capacitor

Also we can see that capacitance is inversely depends on the distance between the plates of capacitor plates

So capacitance will increase if we decrease the distance between two plates

so correct answers are

D) introduce a dielectric material between the plates

E) decrease the separation between the plates

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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:

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where,

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

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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}

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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}

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