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FrozenT [24]
3 years ago
13

An electron is released from rest at a distance of 0.470 m from a large insulating sheet of charge that has uniform surface char

ge density 4.00 × 10^−12 C/m^2 . 1) How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 3.00 × 10^−2 m from the sheet?
Express your answer to three significant figures and include the appropriate units.

2) What is the speed of the electron when it is 3.00 × 10^−2 m from the sheet?
Express your answer to three significant figures and include the appropriate units.
Physics
1 answer:
ArbitrLikvidat [17]3 years ago
7 0

Answer:

Part a)

W = 1.58 \times 10^{-20} J

Part b)

v = 1.86 \times 10^5 m/s

Explanation:

Part a)

Electric field due to large sheet is given as

E = \frac{\sigma}{2\epsilon_0}

\sigma = 4.00 \times 10^{-12} C/m^2

now the electric field is given as

E = \frac{4.00 \times 10^{-12}}{2(8.85 \times 10^{-12})}

E = 0.225 N/C

Now acceleration of an electron due to this electric field is given as

a = \frac{eE}{m}

a = \frac{(1.6 \times 10^{-19})(0.225)}{9.1 \times 10^{-31}}

a = 3.97 \times 10^{10}

Now work done on the electron due to this electric field

W = F.d

d = 0.470 - 0.03

d = 0.44 m

So work done is given as

W = (ma)(0.44)

W = (9.11 \times 10^{-31})(3.97 \times 10^{10})(0.44)

W = 1.58 \times 10^{-20} J

Part b)

Now we know that work done by all forces = change in kinetic energy of the electron

so we will have

W = \frac{1}{2}mv^2 - 0

1.58 \times 10^{-20} = \frac{1}{2}(9.1\times 10^{-31})v^2

v = 1.86 \times 10^5 m/s

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Answer:

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Explanation:

From the question given above, the following data were obtained:

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Next, we shall determine the acceleration of the object. This can be obtained as follow:

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Final velocity (v) = 3.0 m/s

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Answer:

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Answer:

Explanation:

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