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Dafna11 [192]
3 years ago
7

A 3" diameter germanium wafer that is 0.020" thick at 300K has 1.015 x 10^17 As atoms added to it. What is the resistivity of th

e wafer? Germanium has 4.42 x10^22 atoms/cc, electron and hole mobilities are 3900 and 1900 cm^2/(V*s). What is the resistivity of the Ge in ohm*microns?
Physics
1 answer:
Ber [7]3 years ago
7 0

Answer:

0.546 ohm / μm

Explanation:

Given that :

N = 1.015 * 10^17

Electron mobility, u = 3900

Hole mobility, h = 1900

Ng = 4.42 x10^22

q = 1.6*10^-19

Resistivity = 1/qNu

Resistivsity (R) = 1/(1.6*10^-19 * 1.015 * 10^17 * 3900)

= 0.01578880889 ohm /cm

Resistivity of germanium :

R = 1 / 2q * sqrt(Ng) * sqrt(u*h)

R = 1 / 2 * 1.6*10^-19 * sqrt(4.42 x10^22) * sqrt(3900*1900)

R = 1 /0.0001831

R = 5461.4964 ohm /cm

5461.4964 / 10000

0.546 ohm / μm

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OLga [1]

6. "The electric field caused by an electron is weakest near the electron" is FALSE.

7. "An electric field becomes weaker as distance from the electron increase" is TRUE.

<u>Explanation:</u>

The "electrical field" covers the electrical charge and exerts, attracts or repels other charges in the field.The electric field caused by an electron is strongest near the electron while it become weak as distance from the electron increase.

The reason behind is, at a point the direction of the field line is at that point the direction of the field. The relative magnitude of the electric field will be proportional to the field line density. The field is strongest where the field lines are near together and when the field lines are at increasing distance the field is weakest.

5 0
3 years ago
an object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of
Andru [333]

An object is moving with constant velocity downwards on a frictionless inclined plane that makes an angle of θ with the horizontal.

1. Which direction does the force of gravity act on the object?

2. Which direction does the normal force act on the object?

Which force is responsible for the object moving down the incline?

Answer:

The answer is below

Explanation:

1. When an object is moving with a constant velocity, the direction the force of gravity act on the object is DIRECTLY DOWN.

2. When an object is moving with a constant velocity, the direction the normal force act on the object "perpendicular to the surface of the plane."

3. When an object is moving with a constant velocity, the force that is responsible for the object moving down the incline is "the component of the gravitational force parallel to the surface of the inclined plane."

5 0
3 years ago
Point charges q1=+2.00μC and q2=−2.00μC are placed at adjacent corners of a square for which the length of each side is 5.00 cm.
8_murik_8 [283]

The electric potential is a scalar unit, so we don't have to struggle with the vectors. The formula that gives electric potential is

V = \frac{1}{4\pi\epsilon_0}\frac{q}{r}

1) At point a, the electric potential is the sum of the potentials due to q1 and q2. So,

V_a = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

The distance from the center of the square to one of the corners is \sqrt2 L/2 = 0.035m

V_a = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.035} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.035} = 0

The answer is zero, because the point charges are at equal distances and their magnitudes are also equal but their directions are opposite.

2) V_b = \frac{1}{4\pi\epsilon_0}\frac{q_1}{r_1} + \frac{1}{4\pi\epsilon_0}\frac{q_2}{r_2}

r_1 = 0.05\sqrt2m\\r_2 = 0.05m

V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05\sqrt2} + \frac{1}{4\pi\epsilon_0}\frac{-2\times10^{-6}}{0.05}\\V_b = \frac{1}{4\pi\epsilon_0}\frac{2\times10^{-6}}{0.05} (\frac{1}{\sqrt2}-1)\\V_b = \frac{1}{4\pi\epsilon_0} (4\times 10^{-5})(-0.29)\\V_b = (-\frac{2.9\times10^{-6}}{\pi\epsilon_0})[tex]3) The work done on q3 by q1 and q2 is equal to the difference between  energies. This is the work-energy theorem. So,[tex]W = U_b - U_a

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_3}{r} = Vq_3

W = q_3(V_b - V_a) = q_3(V_b - 0)\\W = (-2\times10^{-6})(-\frac{2.9\times10^{-6}}{\pi\epsilon_0})\\W = \frac{5.8\times10^{-12}}{\pi\epsilon_0}

4 0
3 years ago
Whais the cause of necular bomb ​
coldgirl [10]

Answer:

The atomic bomb, is as defined by britannica.com “a deadly weapon caused by the sudden release of energy after the splitting, or fission, of the nuclei of heavy elements like uranium.” In 1945, the United States (US) dropped two atomic bombs, one in Hiroshima and the other in Nagasaki ending WWII.

Explanation:

Hope this helps :)

pls mark brainliest :P

6 0
2 years ago
If the weight of a submerged object is equal to the buoyant force, what net force acts upon the object?
Y_Kistochka [10]

Answer:

A

Explanation:

The weight is acting downwards where as the buoyant force acting upwards (opposite) direction with equal amount of force. so the opposite forces cancel out each other (because of the force amount being equal) and no net force is acting on the object.

Hope i have helped you

Thanks.

7 0
3 years ago
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