Physics, 23.03.2020 06:06 saxondear052

Consider an electron that is 10−10 m from an alpha particle
(q = 3.2 ✕ 10−19 C).
(Enter the magnitudes.)
(a)
What is the electric field (in N/C) due to the alpha particle at the location of the electron?
N/C
(b)
What is the electric field (in N/C) due to the electron at the location of the alpha particle?
N/C
(c)
What is the electric force (in N) on the alpha particle? On the electron?
electric force on alpha particle
N
electric force on electron
N

Answers: 2

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Answers

Answer from: keniaguevara32

a) $2.88\cdot 10^{11} N/C$

b) $1.44\cdot 10^{11} N/C$

c) $4.6\cdot 10^{-8} N$

Explanation:

The magnitude of the electric field produced by a single-point charge is given by the equation

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge source of the field

r is the distance from the charge

The field has direction:

- Out from the charge if the charge is positive

- Towards the charge if the charge is negative

In this part, we want to find the electric field due to the alpha particle at the location of the electron. We have:

$q=3.2\cdot 10^{-19}C$ (charge of the alpha particle)

$r=10^{-10}m$ distance of the electron from the alpha particle

So, the electric field is

$E=\frac{(9\cdot 10^9)(3.2\cdot 10^{-19})}{(10^{-10})^2}=2.88\cdot 10^{11} N/C$

And since the alpha particle is positively charged, the direction of the field is out from the particle.

Here we want to find the electric field due to the electron at the location of the alpha particle.

We use again the same formula as before:

$E=\frac{kQ}{r^2}$

where

k is the Coulomb's constant

Q is the charge source of the field

r is the distance from the charge

where in this case:

$Q=1.6\cdot 10^{-19}C$ is the charge of the electron

$r=10^{-10}m$ distance of the alpha particle from the electron

Therefore, the electric field due to the electron is

$E=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})}{(10^{-10})^2}=1.44\cdot 10^{11} N/C$

And since the electron is negatively charged, the direction of the field is towards the electron.

The electrostatic force between two electric charges is given by Coulomb's Law:

$F=\frac{kq_1 q_2}{r^2}$

where

k is the Coulomb's constant

q1, q2 are the two charges

r is the separation between the two charges

In this problem we have:

$q_1=3.2\cdot 10^{-19}C$ is the charge of the alpha particle

$q_2=1.6\cdot 10^{-19}C$ is the charge of the electron

$r=10^{-10}m$ distance between the two charges

So, the force is

$F=\frac{(9\cdot 10^9)(3.2\cdot 10^{-19})(1.6\cdot 10^{-19})}{(10^{-10})^2}=4.6\cdot 10^{-8} N$

The force is the same on the two particles (Newton's third law of motion), and it is attractive, since the two charges have opposite sign.

Answer from: Quest

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Answer from: Quest

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Consider an electron that is 10−10 m from an alpha particle
(q = 3.2 ✕ 10−19 C).
(Enter...