Question

Protons in an atomic nucleus are typically 10−15 m apart. what is the electric force (in n) of repulsion between nuclear protons? (enter the magnitude.)

Answer 1

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. :

$\large {\boxed {F = k \frac{Q_1Q_2}{R^2} } }$

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:

$F = k \frac{q_1 q_2}{(d)^2}$

$F = k \frac{q^2}{(d)^2}$

$F = 9 \times 10^9 \times \frac{(1.6 \times 10^{-19})^2}{(10^{-15})^2}$

$F = 230.4 \texttt{ Newton}$

$F \approx 230 \texttt{ Newton}$

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Learn more

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

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Answer details

Grade: High School

Subject: Physics

Chapter: Static Electricity

Answer 2

The electric force is given by: 
 F = [ k*(q1)*(q2) ] / d^2 
 F = Electric force 
 k = Coulomb's constant 
 q1 = Charge of one proton 
 q2 = Charge of second proton 
 d = Distance between centers of mass 
 Values: 
 F = unknown 
 k = 8.98E 9 N-m^2/C^2 
 q1 = 1.6E-19 
 q2 = 1.6E-19 
 d = 1.0E-15 m 
 Insert values into F = [ k*(q1)*(q2) ] / d^2 
 F = [ (8.98E 9 N-m^2/C^2) * (1.6E-19) * (1.6E-19) ] / (1.0E-15 m)^2 
 F = 229.888 N
 answer
 the electric force of repulsion between nuclear protons is 229.888 N