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mamaluj [8]
3 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]3 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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An archer puts a 0.30 kg arrow to the bowstring. An average force of 201 N is exerted to draw the string back 1.3 m.a. Assuming
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Explanation:

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(b)if arrow is thrown vertically upward then this energy is converted to Potential energy

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4 0
3 years ago
At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
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Answer:

The thrown rock strike 2.42 seconds earlier.

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This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

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for the dropped rock

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solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

the thrown rock arrives on t=5.4 sec

the dropped rock arrives on t=7.82 sec

so the thrown rock arrives 2.42 seconds earlier (7.82-5.4=2.42)

6 0
3 years ago
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