Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. What is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m

Question

11-09-2019
Physics

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Two protons in an atomic nucleus are typically separated by a distance of 2 ✕ 10-15 m. The electric repulsion force between the protons is huge, but the attractive nuclear force is even stronger and keeps the nucleus from bursting apart. What is the magnitude of the electric force between two protons separated by 2.00 ✕ 10-15 m

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fenixbm fenixbm · Answer 1 · 2019-09-14T04:51:56+02:00

Answer:

The magnitude of the electric force between the to protons will be 57.536 N.

Explanation:

We can use Coulomb's law to find out the force, in scalar form, will be:

[tex]F \ = \ \frac{1}{4 \pi \epsilon_0 } \frac{q_1 q_2}{d^2}[/tex].

Now, making the substitutions

[tex]d \ = \ 2.00 * 10 ^{-15} \ m[/tex],

[tex]q_1 = q_2 = 1.60 * 10 ^ {-19} \ C[/tex],

[tex]\frac{1}{4\pi\epsilon_0}=8.99 * 10^9 \frac{Nm^2}{C^2}[/tex],

we can find:

[tex]F \ = \ 8.99 * 10^9 \frac{Nm^2}{C^2} \frac{(1.60 * 10 ^ {-19} \ C)^2}{(2.00 * 10 ^{-15} \ m)^2}[/tex].

[tex]F \ = 57.536 N[/tex].

Not so big for everyday life, but enormous for subatomic particles.

johanrusli johanrusli · Answer 2 · 2019-12-18T13:08:21+01:00

The magnitude of the electric force between two protons separated by 2.00 × 10⁻¹⁵ m is about 57.6 Newton

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Further explanation

Electric charge consists of two types i.e. positively electric charge and negatively electric charge.

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There was a famous scientist who investigated about this charges. His name is Coulomb and succeeded in formulating the force of attraction or repulsion between two charges i.e. :

[tex]\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }[/tex]

F = electric force (N)

k = electric constant (N m² / C²)

q = electric charge (C)

r = distance between charges (m)

The value of k in a vacuum = 9 x 10⁹ (N m² / C²)

Let's tackle the problem now !

[tex]\texttt{ }[/tex]

Given:

distance between protons = d = 2 × 10⁻¹⁵ m

charge of proton = q = 1.6 × 10⁻¹⁹ C

Unknown:

electric force = F = ?

Solution:

[tex]F = k \frac{q_1 q_2}{(d)^2}[/tex]

[tex]F = k \frac{q^2}{(d)^2}[/tex]

[tex]F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(2 \times 10^{-15})^2}[/tex]

[tex]F = 57.6 \texttt{ Newton}[/tex]

[tex]\texttt{ }[/tex]

Learn more

The three resistors : https://brainly.com/question/9503202
A series circuit : https://brainly.com/question/1518810
Compare and contrast a series and parallel circuit : https://brainly.com/question/539204

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Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity