The electric force of repulsion between nuclear protons is about 230 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 !
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Given:
distance between protons = d = 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}{(10^{-15})^2}[/tex]
[tex]F = 230.4 \texttt{ Newton}[/tex]
[tex]F \approx 230 \texttt{ Newton}[/tex]
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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
