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mamaluj [8]
2 years ago
14

An electron enters the gap between the plates of a capacitor at the center of the gap traveling parallel to theplates at 2.0 x 1

06m/s parallel to the plates. If the gap has a width of 1.0 cm and the surface charge densityof the upper and lower plates are±1.0 x 10−6C/m2, respectively. How far will the electron travel beforehitting a plate? (Assume that the electric field is uniform inside the capacitor and zero outside.)
Physics
1 answer:
Svetlanka [38]2 years ago
4 0

Answer:

How far will the electron travel beforehitting a plate is 248.125mm

Explanation:

Applying Gauss' law:

Electric Field E = Charge density/epsilon nought

Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12

Therefore E = 1.0 x 10^-6/8.85× 10^-12

E= 1.13×10^5N/C

Force on electron F=qE

Where q=charge of electron=1.6×10^-19C

Therefore F=1.6×10^-19×1.13×10^5

F=1.808×10^-14N

Acceleration on electron a = Force/Mass

Where Mass of electron = 9.10938356 × 10^-31

Therefore a= 1.808×10^-14 /9.11 × 10-31

a= 1.985×10^16m/s^2

Time spent between plate = Distance/Speed

From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2

Therefore Time = 0.01/2×10^6

Time =5×10^-9s

How far the electron would travel S =ut+ at^2/2 where u=0

S= 1.985×10^16×(5×10^-9)^2/2

S=24.8125×10^-2m

S=248.125mm

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HACTEHA [7]

\text{Given that,}\\\\\text{Initial velocity,} ~v_0 = 0~ \text{m~s}^{-1}\\\\\text{Time,  t = 1.3~sec}\\\\\text{Acceleration, a = 9.1 m s}^{-2}\\\\\\\\\text{Velocity,}\\\\v = v_0  +at\\\\\implies v = 0 + 9.1 \times 1.3 = 11.83~~ \text{m~s}^{-1}

5 0
2 years ago
5. An acrobat, starting from rest, swings freely on a trapeze of
34kurt

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

            Em₀ = U = m g h

Final point. Lower part of the movement

            Em_f = K = ½ m v²

Energy is conserved.

            Emo = Em_f  

            m g h = ½ m v²

            v² = 2 gh

Let's use trigonometry to find the height, see attached.

         h = L - L cos θ

         h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

         h = 3.7 (1- cos 48)

          h = 1.22 m

this  is the maximum height of the movement.

Let's calculate the velocity.  

          v= \sqrt{2 \ 9.8 \ 1.22}  

          v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

Learn more here: brainly.com/question/13010190

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2 years ago
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lisov135 [29]

Answer:

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Explanation: hope this helps :)

7 0
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Newton’s third law of motion says that for every action there is a(n) and opposite reaction.
sattari [20]

Answer:

for every action thete is an equal and opposite reaction

3 0
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An athlete practicing for a track meet ran 400 meter in 50 seconds. What was his average speed?
sattari [20]

Speed= Distance ÷ time

Speed= 400 ÷ 50 =8 m/s

7 0
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