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3 years ago

5

An electron is placed 5.40 cm due north of a charged sphere and experienced a force of 8.30 x 10^-13 N due north. What is the el

ectric field experienced by the electron at that point?

1 answer:

ira [324]3 years ago

4 0

Answer:

The electric field is $E = 5.1*10^6N/C.$

Explanation:

The force $F$ on a charge $q$ in an electric field $E$ is given by

$F = qE$ ,

which can be rearranged to give

$E = \dfrac{F}{q }$

Now, the force on the electron is $F = 8.30*10^{-13}N$ , and its charge is

$q = 1.6*10^{-19}C$ ; therefore,

$E = \dfrac{8.30*10^{-13}N}{1.6*10^{-19}C}$

$\boxed{E = 5.1*10^6N/C}$

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Answer:

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Explanation:

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here we know that

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3 years ago

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Answer:

B

Explanation:

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3 years ago

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Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

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so we will have

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now we know that time to cross the plates is given as

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3 years ago

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soldi70 [24.7K]

Answer:

98.13m

Explanation:

Complete question

Daniel is 50.0 meters away from a building. Tip of the building makes an angle of 63.0° with the horizontal. What is the height of the building

CHECK THE ATTACHMENT

From the figure, using trigonometry

Tan(θ ) = opposite/adjacent

Where Angle (θ )= 63°

Opposite= X = height of the building

Adjacent= 50 m

Then substitute the values we have

Tan(63)= X/50

1.9626= X/50

X= 1.9626 × 50

X= 98.13m

Hence, the height of the building is 98.13m

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Answer:

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Explanation:

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3 years ago