# The electrons in a particle beam each have a kinetic energy of 1.60×10−17 J. What are the magnitude and direction of the electric ﬁeld that will stop these electrons in a distance of 10.0cm?

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To find the magnitude and direction of the electric field that will stop the electrons in a distance of 10.0 cm, we can use the formula:

[ K = q \cdot V ]

Where:

- ( K ) is the kinetic energy of the electron,
- ( q ) is the charge of the electron,
- ( V ) is the potential difference.

Given that the kinetic energy of each electron is ( 1.60 \times 10^{-17} ) J and the distance is 10.0 cm, we can calculate the potential difference using the formula:

[ K = q \cdot V ]

[ V = \frac{K}{q} ]

Given that the charge of an electron ( q ) is ( 1.6 \times 10^{-19} ) C, we can calculate ( V ).

[ V = \frac{1.60 \times 10^{-17}}{1.6 \times 10^{-19}} ]

[ V = 100 , \text{V} ]

Now, to find the magnitude of the electric field (( E )), we use the formula:

[ E = \frac{V}{d} ]

Where ( d ) is the distance traveled by the electron.

[ E = \frac{100}{0.1} ]

[ E = 1000 , \text{V/m} ]

The magnitude of the electric field required to stop the electrons in a distance of 10.0 cm is ( 1000 , \text{V/m} ). Since the electric field is directed opposite to the direction of the electron's motion, its direction is opposite to the initial direction of the electron beam.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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